C, C++, and Assembly Language Programs & Tutotrials

Add/Sum two numbers using pointers in C/C++ Language

noreply@blogger.com (Maaz Ali) — Mon, 13 Nov 2017 19:56:00 +0000

Problem

Take two integer inputs from user. Create Sum function which should receive two integers using pointers. Calculate the sum and display the result in main function.

Code

#include "stdafx.h"
#include <iostream>
#include<conio.h>
using namespace std;

int sum (int *, int *);

int main()
{
int a,b,c;

cout << "Enter 1st number: ";
cin >> a;
cout<<"enter 2nd number: ";
cin >> b;

c = sum (&a,&b);

cout << "Sum = "<< c << endl;

cout<<"Press any key to close.";

getch();

return 0;
}

int sum (int *a, int *b)
{
int c;
c = (*a + *b);
return c;
}

Output

Add/Sum two numbers using pointers in C/C++ Language

Calculate Square of 3 input integers using function and pointers

noreply@blogger.com (Maaz Ali) — Sat, 04 Nov 2017 09:34:00 +0000

Problem

Write a program to calculate square of an integer, input by the user. Declare a function with the name of square and should receive 3 integers. The function should then calculate the square of these three integers and return the square. Display the result.

Code

#include "stdafx.h"
#include <iostream>
#include<conio.h>
using namespace std;

void square (int *, int *, int *);

int main()
{
int a,b,c;

cout << "Enter 1st number: ";
cin >> a;

cout << "Enter 2nd number: ";
cin >> b;

cout << "Enter 3rd number: ";
cin >> c;

square (&a, &b, &c);

cout << "Square of Ist number = "<< a;
cout << "\nSquare of 2nd number = "<< b;
cout << "\nSquare of 3rd number = "<< c << endl ;

cout<<"Press any key to close.";

getch();

return 0;
}

void square (int *a, int *b, int *c)
{
*a = (*a * *a);
*b = (*b * *b);
*c = (*c * *c);
}

Output

Calculate Square of 3 input integers using function and pointers

Display Add/Sum of two integers in C Language

noreply@blogger.com (Maaz Ali) — Sun, 15 Oct 2017 18:33:00 +0000

Problem

Write a program in C language which takes two integers as input from user. Write a function to add these two integers and store the sum in another integer. Display the result.

Code

// lab02..cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
#include<conio.h>
using namespace std;

int sum (int, int);

int main()
{
int a,b,c;

cout << "Enter 1st number: ";
cin >> a;
cout<<"enter 2nd number: ";
cin >> b;

c = sum (a,b);

cout << "Sum = "<< c << endl;

cout<<"Press any key to close.";

getch();

return 0;
}

int sum (int a, int b)
{
int c;
c = a+b;
return c;
}

Output

Display Add/Sum of two integers in C Language

Date Sum Using Structure in C/C++ Language

noreply@blogger.com (Maaz Ali) — Sat, 07 Oct 2017 19:28:00 +0000

Problem

Create a struct called data. User should be able to input day, month, and year. Take two date inputs and add them. Display the result.

Code

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<string.h>

//struct decleration
struct data {
int years;
int months;
int days;
}d1,d2;

//function protorypes
void add_data();
void add_dates();
void display(int ,int ,int);

int main()
{
//function calls
add_data();
add_dates();

printf ("\n\nPress any key to exit.");
getch();
return 0;
}

//adding record
void add_data()
{
//1st date
printf ("Enter 1st date.");

printf ("\n\nEnter year: ");
scanf ("%d",&d1.years);

printf ("\nEnter month: ");
scanf ("%d",&d1.months);

printf ("\nEnter day: ");
scanf ("%d",&d1.days);

display(d1.days,d1.months,d1.years);
//printf ("\n\nThe 1st date is %d/%d/%d",d1.days,d1.months,d1.years);

//Second time
printf ("\n\n\nNow, please type the 2nd date.");

printf ("\nEnter year: ");
scanf ("%d",&d2.years);

printf ("\nEnter month: ");
scanf ("%d",&d2.months);

printf ("\nEnter day: ");
scanf ("%d",&d2.days);

display(d2.days,d2.months,d2.years);
}

void display (int a,int b,int c)
{
printf("\nThe date is %d/%d/%d \n\n", a,b,c);
}

void add_dates()
{
int y,m,d;

y = d1.years + d2.years;
m = d1.months + d2.months;
d = d1.days + d2.days;

printf ("\n\n\nSum of the date's is %d/%d/%d",d,m,y);
}

Output

Date Sum Using Structure in C/C++ Language

Time Sum Using Structure in C/C++ Language

noreply@blogger.com (Maaz Ali) — Sun, 01 Oct 2017 09:08:00 +0000

Problem

Create a structure called time. Its three members, all type int, should be called hours, minutes, and seconds. Write a program that prompts the user to enter a time value in hours, minutes, and seconds. This can be in 12:59:59 format, or each number can be entered at a separate prompt (“Enter hours:”, and so forth). The program should then store the time in a variable of type struct time, and finally print out the total number of seconds represented by this time value:

long totalsecs = t1.hours*3600 + t1.minutes*60 + t1.seconds

Code

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<string.h>

struct time {
int hours;
int minutes;
int seconds;
};

//function prototype
void add (int,int,int,int,int,int);

int main()
{
struct time t1,t2;

//1st time
printf ("Enter 1st time.");

printf ("\n\nEnter Hours: ");
scanf ("%d",&t1.hours);

printf ("\nEnter Minutes: ");
scanf ("%d",&t1.minutes);

printf ("\nEnter Seconds: ");
scanf ("%d",&t1.seconds);

printf ("\n\nThe Time is %d:%d:%d",t1.hours,t1.minutes,t1.seconds);

//Second time
printf ("\n\n\nNow, please type the 2nd time.");

printf ("\n\nEnter Hours: ");
scanf ("%d",&t2.hours);

printf ("\nEnter Minutes: ");
scanf ("%d",&t2.minutes);

printf ("\nEnter Seconds: ");
scanf ("%d",&t2.seconds);

printf ("\n\nThe Time is %d:%d:%d",t2.hours,t2.minutes,t2.seconds);

//function call
add (t1.hours,t1.minutes,t1.seconds,t2.hours,t2.minutes,t2.seconds);

printf ("\n\nPress any key to exit.");
getch();
return 0;
}

//function definition
void add (int x,int y,int z,int a,int b,int c)
{
int h,m,s;

h = x + a;
m = y + b;
s = z + c;

printf ("\n\n\nSum of the two time's is %d:%d:%d",h,m,s);
}

Code

Time Sum Using Structure in C/C++ Language

Input Length and Breadth. Calculate Area and Perimeter Using Struct in C Language

noreply@blogger.com (Maaz Ali) — Sun, 24 Sep 2017 13:00:00 +0000

Problem:

Write a program which takes length and breadth as input. Than calculate area and perimeter of the room.

Code

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<string.h>

struct distance{
int length;
int width;
};
struct room{
struct distance a;
}r;

void myftn();

int main()
{
myftn();

printf ("\n\nPress any key to exit.");
getch();
return 0;
}

void myftn ()
{
int p,s;

printf ("Enter lengtgh: ");
scanf ("%d",&r.a.length);
printf ("Enter widtgh: ");
scanf ("%d",&r.a.width);

p = 2 * (r.a.length * r.a.width);
s = (r.a.length * r.a.width);

printf ("\n\nArea = %d",s);
printf ("\nPerimeter = %d",p);
}

Output

Input Length and Breadth. Calculate Area and Perimeter Using Struct in C Language

Convert Fahrenheit to Celsius | Convert Celsius to Fahrenheit

noreply@blogger.com (Maaz Ali) — Sun, 17 Sep 2017 14:50:00 +0000

Problem

Write two functions in a program of C language to convert:

Function 1: Fahrenheit to Celsius
Function 2: Celsius to Fahrenheit

Code

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<string.h>

void ftoc();

void ctof();

int main()

{

ftoc();

ctof();

printf ("\n\nPress any key to exit.");

getch();

return 0;

}

void ftoc()

{

float f,c;

printf ("Enter temperature in Fahrenheit: ");

scanf ("%f",&f);

c = 5.0 / 9.0 * (f-32.0);

printf ("Temperature in celsius: %f",c);

}

void ctof()

{

float f,c;

printf ("\n\nEnter temperature in celsius: ");

scanf ("%f",&c);

f = ((9.0 / 5.0 * c) + 32.0);

printf (" Temperature in celsius: %f",f);

}

Output

Convert Fahrenheit to Celsius | Convert Celsius to Fahrenheit

File Handling: Reading and Writing into a text file using class

noreply@blogger.com (Maaz Ali) — Thu, 27 Apr 2017 11:35:00 +0000

Code

#include<iostream>
#include<fstream>
#include<string>
#include<conio.h>
using namespace std;

class marks
{
private:
int oop;
int cal;
int dis;
int s;
int a;

public:
marks():oop(0),cal(0),dis(0)
{ }

void getdata()
{
cout<<"Enter marks of oop: ";
cin>>oop;
cout<<"Enter marks of cal: ";
cin>>cal;
cout<<"Enter marks of dis: ";
cin>>dis;
}

void display()
{
cout<<"OOP = "<<oop<<endl;
cout<<"Cal = "<<cal<<endl;
cout<<"Dis = "<<dis<<endl;
}
};

int main ()
{
marks m;
int s;

m.getdata();
//opening the file in binary mode
ofstream myfile("Writing_Reading_class.txt",ios::binary);

//writing in the file
myfile.write(reinterpret_cast<char *>(&m),sizeof(m));
//closing the file
myfile.close();
//re-opening the file in binary mode
ifstream my("Writing_Reading.txt",ios::binary);
//Reading from the file
my.read(reinterpret_cast<char *>(&m),sizeof(m));
//display
m.display();

cout << "\n\nPress any key to close.";
getch ();
return 0;
}

Output

File Handling: Reading and Writing into a binary text file

noreply@blogger.com (Maaz Ali) — Sat, 15 Apr 2017 06:19:00 +0000

Code:

#include<iostream>
#include<fstream>
#include<string>
#include<conio.h>
using namespace std;

int main()
{
int max = 10;
int a[max];

ofstream myfile("Writing_Reading.txt",ios::binary); //opening the file in binary mode

for(int j=0;j<max;j++) //assigning values(array)
a[j] = (j + 1);

//writing in the file
myfile.write(reinterpret_cast<char *>(a),sizeof(a));
cout<<"File written."<<endl<<endl;

myfile.close(); //closing the file

ifstream my("Writing_Reading.txt",ios::binary); //opening the file in binary mode

//Emptying the array
for(int j=0;j<max;j++)
a[j] = 0;
//Reading from the file
my.read(reinterpret_cast<char *>(a),sizeof(a));

for(int j=0;j<max;j++) //display array
cout<<a[j];

cout<<endl<<"Press any key to exit.";
getch();
return 0;
}

Output:

find index number of the largest element using class

noreply@blogger.com (Maaz Ali) — Fri, 10 Mar 2017 14:39:00 +0000

Problem

Start with a program that allows the user to input a number of integers, and then stores them in an int array. Write a function called maxint() that goes through the array, element by element, looking for the largest one. The function should take as arguments the address of the array and the number of elements in it, and return the index number of the largest element. The program should call this function and then display the largest element and its index number.

Code

// p[4].cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include<conio.h>
#include<iostream>
#include<string.h>
using namespace std;

class intgrs
{
private:
int a[5];
int i,j,z;

public:
intgrs():j(0),i(0),z(0)
{ }

void maxint()
{
cout<<"Enter Integers: ";
for(i=0; i<5; i++)
{
cin>>a[i];
}
for(i=0;i<5;i++)
{
if(a[i]>j)
{
j = a[i];
z=i;
}
}
cout<<"\n\nLargest Element = "<<j;
cout<<"\n\nIndex Number = "<<(z+1)<<endl;
}
};

int main()
{
intgrs i;

i.maxint();

cout<<"\n\nPress any key to close.";
getch();
return 0;
}

Output

Input and display employee data of 100 employees using array of objects

noreply@blogger.com (Maaz Ali) — Thu, 09 Mar 2017 15:04:00 +0000

Problem

Create a class called employee that contains a name (an object of class string) and an employee number (type long). Include a member function called getdata() to get data from the user for insertion into the object, and another function called putdata() to display the data. Assume the name has no embedded blanks.
Write a main() program to exercise this class. It should create an array of type employee, and then invite the user to input data for up to 100 employees. Finally, it should print out the data for all the employees.

Code

#include "stdafx.h"
#include<conio.h>
#include<iostream>
#include<string.h>
using namespace std;

class employee
{
private:
char name[50];
long no;

public:
employee():no(0)
{ }

void getdata()
{
fflush(stdin);
cout<<"\nEnter Employee Data:\n";
cout<<"Enter Name: ";
cin.get(name,50);
fflush(stdin);
cout<<"Enter Number: ";
cin>>no;
fflush(stdin);
}

void putdata()
{
cout<<"\n\nName= "<<name<<"\nE.number = "<<no<<endl;
}

};

int main()
{
int array_length = 3;
employee e1[array_length];

for(int i=0;i<array_length;i++){
e1[i].getdata();
fflush(stdin);
}

for(int i=0;i<array_length;i++){
e1[i].putdata();
}

cout<<"\n\nPress any key to close.";
getch();
return 0;
}

Output

Reverse input string in c++

noreply@blogger.com (Maaz Ali) — Sat, 22 Oct 2016 12:08:00 +0000

Problem

Take a string input from user and reverse it using C++ language.

Code

// P[1].cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include<conio.h>
#include<iostream>
#include<string.h>
using namespace std;

class rvrstr
{
private:
char str1[80];
char str2[80];
int i,j,n;
char ch;

public:
rvrstr(): i(0),j(0),n(0)
{ }

void setstr()
{
cout<<"Enter string: ";
cin.get(str1,80);
}

void reversit()
{
n = strlen(str1);
cout<<"\n\nReversed String: ";
for(i=n;i>=0;i--)
{
str2[j] = str1[i];
cout<<str2[j];
j++;
}
}
};

int main()
{
rvrstr r1;

r1.setstr();
r1.reversit();

cout<<"\n\nPress any key to close.";
getch();
return 0;
}

Output

Electronic Polling System in C++ Language

noreply@blogger.com (Maaz Ali) — Thu, 20 Oct 2016 09:01:00 +0000

Problem

Implement (a part of) electronic Pakistani election Scenario where following political parties are participating in election: PPP, PML and PTI. Create a class PollingStation where it stores the vote’s count, against each party, cast by the Pakistani legal registered voters. Your program must query the voter for its ID card number before he/she cast the vote and check it against voter list. (Suppose voter list is already with you). If voter is not found in the voter’s list, your program throws the exception IllegalVoter() and discard that voter. The Application again starts working in normal state and terminates only when user wants to quit it. At the end of the day (or whenever user wants) it generates report about the vote count against each party including header that contains date, day and location.

Code

// Poling station.cpp : Defines the entry point for the console application.
//
//HEADER FILES
#include "stdafx.h" //ESSENTIAL TO WORK ON VISUAL STUDIO
#include<fstream> //FOR FILE
#include<conio.h> //FOR GETCH FUNCTION
#include<string> //fOR STRINGS USAGE
#include<iostream> //FOR COUT AND CIN
using namespace std;
//CLASS BEGINS
class PollingStation
{
private: //VARIABLES DECLARED PRIVATLY, ONLY ACCESSABLE IN THIS CLASS
int votes_count_ppp;
int votes_count_pml;
int votes_count_pti;
int total_votes,hr,mn,sc,flag,choice,offset;
string line,day,location;

public: //EXCEPTION CLASS
class IllegalVoter
{ };
//CONSTRUCTOR OF POLLING STATION CLASS
PollingStation(): votes_count_ppp(0), votes_count_pml(0), votes_count_pti(0),total_votes(0),
hr(0),mn(0),sc(0),flag(0),line(""),choice(0)
{ }
//DESTRUCTOR OF POLLING STATION CLASS
~PollingStation()
{
cout<<"polling Station Destructor";
}
//GETING THE DATA FROM USER(DAY,DTAE & LOCATION)
void start()
{ //GET DAY
system("cls");
cout<<"Enter Day: ";
cin>>day;
//GET LOCATION
system("cls");
cout<<"\nEnter Location: ";
cin>>location;
//GET TIME
system("cls");
cout<<"\nEnter time: "<<endl;
cout<<"Enter Hour: ";
cin>>hr;
cout<<"Enter Minutes: ";
cin>>mn;
}
//CHECKING WHETHER THE PERSON IS ILLEGIBEL FOR VOTE
void checkdata(string search)
{
ifstream Myfile; //OPENING THE FILE FOR READING
Myfile.open ("VOTERS_LIST.txt");
{
if(Myfile.is_open())
{ //READING UNTIL END OF FILE
while(!Myfile.eof())
{
getline(Myfile,line); //CHECKIMG
if ((offset = line.find(search)) != string::npos)
{
flag = 1;
}
}
Myfile.close();
} //IF FILE IS NOT OPENED
else
cout<<"Unable to open this file."<<endl;
}
if(flag == 0) //THROWING EXCEPTION
throw IllegalVoter();
}
//PERSON WILL CAST HIS/HER VOTE
void cast_vote()
{
system("cls");
cout<<"CAST YOUR VOTE:"<<endl;
cout<<"1. PPP"<<endl;
cout<<"2. PML"<<endl;
cout<<"3. PTI"<<endl;

cout<<"\nEnter Your Choice: ";
cin>>choice;
//IF USER HAS VOTED FOR PPP
if(choice == 1)
{
votes_count_ppp++;
total_votes++;
} //IF USER HAS VOTED FOR PML
else if(choice == 2)
{
votes_count_pml++;
total_votes++;
} //IF USER HAS VOTED FOR PTI
else if (choice == 3)
{
votes_count_pti++;
total_votes++;
} //IF USER HAS ENTERED WRONG CHOICE
else
{
system("cls");
cout<<"Invalid Key."<<endl;
cout<<"Press Any Key To Continue."<<endl;
getch();
}
}
//DISPLYING RESULTS
void display()
{
system("cls");
cout<<"\n\n********************************************************************************";
cout<<"Time = "<<hr<<" : "<<mn<<"\t\t\tDay = "<<day<<"\t\tLocation = "<<location<<endl;
cout<<"\n********************************************************************************";

cout<<"Total number of votes voted for:"<<endl<<endl;
cout<<"\tPPP = "<<votes_count_ppp<<endl;
cout<<"\tPML = "<<votes_count_pml<<endl;
cout<<"\tPTI = "<<votes_count_pti<<endl<<endl;
cout<<"\tTotal = "<<total_votes<<endl;
}
};

int main()
{
PollingStation p; //OBJECT OF POLLING STATION CLASS
char another; //VARIABLES
string id;

p.start(); //CALLING THE FUNTION TO GET DATA

do{ //LOOP UNTIL USER WANTS
try{
system("cls"); //USER WILL ENTER HIS/HER ID CARD NUMBER
cout<<"Enter ID Card Number: ";
cin>>id;
p.checkdata(id); //CHECKING
p.cast_vote(); //CASTING VOTE(IF ILLEGIBLE)
}
catch(PollingStation::IllegalVoter) //IF NOT ILLEGIBLE THEN
{
cout<<"NO ENTRY WITH THIS ID"<<endl;
cout<<"Press any key to continue...";
getch();
}

system("cls"); //ASKING USER WHETHER HE/SHE WANTS TO ENTER DATA
cout<<"Enter another phone number(y/n)? ";
another = getch();
}while(another == 'Y' || another == 'y'); //LOOP TERMINATION CONDITION

ofstream file; //OPENING FILE FOR WRITTING
file.open("POLLING_RESULTS.DAT", ios::binary);
p.display(); //DISPLYING THE DATA & WRITTING TO THE FILE
file.write(reinterpret_cast<char*>(&p),sizeof(p)); //SO THAT IT COULD BE DSPLAYED ANYTIME
file.close(); //USER WANTS & THEN CLOSING THE FILE

cout<<"\nFile Written"<<endl;
cout<<"Press any key to exit"<<endl;
getch();
return 0;
}

Output

Polymorphism example in c++ language

noreply@blogger.com (Maaz Ali) — Wed, 19 Oct 2016 14:45:00 +0000

Problem

Implement the following given scenario, include two constructors (no argument, with arguments), two functions getvalue(), and displayvalue() for all classes.

Code

#include<iostream>
#include<conio.h>
#include<string>
using namespace std;

class Vehicle
{
protected:
int i;
string s;

public:
Vehicle(): i(0), s("")
{ }
Vehicle(int a,string ss): i(a), s(ss)
{ }
void getdata()
{
cout<<"Enter Vehicle Number: ";
cin>>i;
fflush(stdin);
cout<<"Enter vehicle name: ";
getline(cin, s);
}

void display()
{
cout<<"Vehicle Number: "<<i;
cout<<"Vehicle Name: "<<s;
}
};

class land_Vehicle : public Vehicle
{
protected:
int b;
string s1;

public:
land_Vehicle(): b(0), s1("")
{ }
land_Vehicle(int a,string ss, int r, string s0): b(a), s1(ss), Vehicle(r,s0)
{ }
void getdata()
{
Vehicle :: getdata();
cout<<"Enter Land Vehicle Number: ";
cin>>b;
fflush(stdin);
cout<<"Enter Land vehicle name: ";
getline(cin, s1);
}

void display()
{
Vehicle :: display();
cout<<"Vehicle Number: "<<b;
cout<<"Vehicle Name: "<<s1;
}
};

class Boat : public Vehicle
{
protected:
int c;
string s2;

public:
Boat(): c(0), s2("")
{ }
Boat(int a,string ss, int r, string s0): c(a), s2(ss), Vehicle(r,s0)
{ }
void getdata()
{
Vehicle :: getdata();
cout<<"Enter Boat Number: ";
cin>>c;
fflush(stdin);
cout<<"Enter Boat name: ";
getline(cin, s2);
}

void display()
{
Vehicle :: display();
cout<<"Vehicle Number: "<<c;
cout<<"Vehicle Name: "<<s2;
}
};

class aircraft : public Vehicle
{
protected:
int d;
string s3;

public:
aircraft(): d(0), s3("")
{ }
aircraft(int a,string ss, int r, string s0): d(a), s3(ss), Vehicle(r,s0)
{ }
void getdata()
{
Vehicle :: getdata();
cout<<"Enter Aircraft Number: ";
cin>>d;
fflush(stdin);
cout<<"Enter Aircarft name: ";
getline(cin, s3);
}

void display()
{
Vehicle :: display();
cout<<"Vehicle Number: "<<d;
cout<<"Vehicle Name: "<<s3;
}
};

class car : public land_Vehicle
{
protected:
int e;
string s4;

public:
car(): e(0), s4("")
{ }
car(int a,string ss, int q1, string q2,int q3,string q4): e(a), s4(ss),land_Vehicle(q1,q2,q3,q4)
{
}
void getdata()
{
land_Vehicle :: getdata();
cout<<"Enter car Number: ";
cin>>e;
fflush(stdin);
cout<<"Enter car name: ";
getline(cin, s4);
}

void display()
{
land_Vehicle :: display();
cout<<"Vehicle Number: "<<e;
cout<<"Vehicle Name: "<<s4;
}
};

class truck : public land_Vehicle
{
protected:
int f;
string s5;

public:
truck(): f(0), s5("")
{ }
truck(int a,string ss, int q, string qq,int q3,string q4): f(a), s5(ss),land_Vehicle(q,qq,q3,q4)
{ }
void getdata()
{
land_Vehicle :: getdata();
cout<<"Enter truck Number: ";
cin>>f;
fflush(stdin);
cout<<"Enter truck name: ";
getline(cin, s5);
}

void display()
{
land_Vehicle :: display();
cout<<"Vehicle Number: "<<f;
cout<<"Vehicle Name: "<<s5;
}
};

class sports_car : public car
{
protected:
int g;
string s6;

public:
sports_car(): g(0), s6("")
{ }

sports_car(int a,string ss, int q, string qq,int q3,string q4,int q5,string q6): g(a), s6(ss),car(q,qq,q3,q4,q5,q6)
{ }
void getdata()
{
car:: getdata();
cout<<"Enter sports car Number: ";
cin>>g;
fflush(stdin);
cout<<"Enter sports car name: ";
getline(cin, s6);
}

void display()
{
car:: display();
cout<<"Vehicle Number: "<<g;
cout<<"Vehicle Name: "<<s6;
}
};

class yacht : public Boat
{
protected:
int h;
string s7;

public:
yacht(): h(0), s7("")
{ }
yacht(int a,string ss, int q, string qq,int q3,string q4): h(a), s7(ss),Boat(q,qq,q3,q4)
{ }
void getdata()
{
Boat::getdata();
cout<<"Enter yacht Number: ";
cin>>h;
fflush(stdin);
cout<<"Enter yacht name: ";
getline(cin, s7);
}

void display()
{
Boat :: display();
cout<<"Vehicle Number: "<<h;
cout<<"Vehicle Name: "<<s7;
}
};

class sail_boat : public Boat
{
protected:
int k;
string s8;

public:
sail_boat(): k(0), s8("")
{ }
sail_boat(int a,string ss, int q, string qq,int q3,string q4): k(a), s8(ss),Boat(q,qq,q3,q4)
{ }
void getdata()
{
Boat :: getdata();
cout<<"Enter sail_boat Number: ";
cin>>k;
fflush(stdin);
cout<<"Enter sail_boat name: ";
getline(cin, s8);
}

void display()
{
Boat :: display();
cout<<"Vehicle Number: "<<k;
cout<<"Vehicle Name: "<<s8;
}
};

class f16 : public aircraft
{
protected:
int j;
string s9;

public:
f16(): j(0), s9("")
{ }
f16(int a,string ss, int q, string qq,int q3,string q4): j(a), s9(ss), aircraft(q,qq,q3,q4)
{ }
void getdata()
{
aircraft :: getdata();
cout<<"Enter f16 Number: ";
cin>>j;fflush(stdin);
cout<<"Enter f16 model: ";
getline(cin, s9);
}

void display()
{
aircraft :: display();
cout<<"f16 Number: "<<j;
cout<<"f16 model: "<<s9;
}
};

class helicopter : public aircraft
{
protected:
int l;
string s10;

public:
helicopter(): l(0), s10("")
{ }
helicopter(int a,string ss, int q, string qq,int q3,string q4): l(a), s10(ss),aircraft(q,qq,q3,q4)
{ }
void getdata()
{
aircraft :: getdata();
cout<<"Enter helicopter Number: ";
cin>>l;
cout<<"Enter helicopter model: ";
getline(cin, s10);
}

void display()
{
aircraft :: display();
cout<<"helicopter Number: "<<l;
cout<<"helicopter model: "<<s10;
}
};

int main()
{
aircraft a;
a.getdata();
a.display();

Boat b;
b.getdata();
b.display();

land_Vehicle l;
l.getdata();
l.display();

car c;
c.getdata();
c.display();

truck t;
t.getdata();
t.display();

sports_car sp;
sp.getdata();
sp.display();

yacht y;
y.getdata();
y.display();

sail_boat s;
s.getdata();
s.display();

f16 f;
f.getdata();
f.display();

helicopter h;
h.getdata();
h.display();

cout<<"Press any key exit.";
getch();
return 0;
}

Output

CD player simulator using c++

noreply@blogger.com (Maaz Ali) — Fri, 14 Oct 2016 06:07:00 +0000

Problem

Design a simple CD player simulator. The CD player has the following buttons: "POWER", "EJECT", "PLAY" and "STOP", and a tray where a CD is loaded. User can load a CD only if the tray is open. The tray can be opened only if the power is turned on (regard it as something like a motor-driven tray installed on a PC). If the power is turned off with the tray open, the tray will keep open and the power is turned off.
Implement the following actions of the CD player as methods of class CDPlayer.

pushPower() -- simulates pushing POWER(on/off) button.

if the power is turned off, turn on the power, and display "Power turned on."
if the power is turned on,

if playing, stop it.

turn off the power, and display "Power turned off."

pushEject() -- simulates pushing EJECT(tray open/close) button

if the power is not turned on, display "Power not turned on."
if the power is turned on,

if the tray is not open, open the tray and display "Tray opened."
if the tray is open, close the tray and display "Tray closed."

pushPlay() -- simulates pushing PLAY button

if the power is turned on,

if the media is loaded and tray is closed, begin to play and display "Now playing..."
otherwise, display "No CD loaded or tray is open."

pushStop() -- simulates pushing STOP button

if the power is turned on,

if it's playing, stop playing and display "Playing stopped."

loadUnloadMedia() -- simulates loading/unloading media to/from the tray (tray must be open when loading/unloading)

if the tray is open,

if the media is loaded, unload the media and display "Media unloaded."
if the media is not loaded, load the media and display "Media loaded."

if the tray is not open,

display "Tray is not open."

Code

#include<iostream>

#include<conio.h>

using namespace std;

class CDplayer

{

private:

int power;

int eject;

int tray;

int media;

int play;

public:

CDplayer():power(0),eject(0),tray(0),media(0),play(0)

{

}

void pushPower()

{

cout<<"\n\nNOTE: Press one(1) to turn on and zero(0) to turn off.";

cout<<"\nWant to turn on the power? ";

cin>>power;

if(power == 1)

cout<<"Power is ON.";

else if (power == 0)

cout<<"Power is OFF.";

else

cout<<"Invalid key";

cout<<"\n\nPress any key to continue.";

getch();

}

void pushEject()

{

if(power == 1)

{

if(tray == 0)

{

tray = 1;

cout<<"Tray opened.";

}

else if(tray == 1)

{

tray = 0;

cout<<"Tray closed.";

}

else if(power == 0)

cout<<"Power not turned on.";

cout<<"\n\nPress any key to continue.";

getch();

}

void pushPlay()

{

if(power == 1)

{

if(media == 1 && tray == 0)

{

play = 1;

cout<<"Now playing...";

}

else

cout<<"No CD loaded or tray is open.";

}

else if(power == 0)

cout<<"Power not turned on.";

cout<<"\n\nPress any key to continue.";

getch();

}

void pushStop()

{

if(power == 1)

{

if(play == 1)

{

play = 0;

cout<<"Playing stopped.";

}

cout<<"\n\nPress any key to continue.";

getch();

}

void loadUnloadMedia()

{

if(tray == 1)

{

if(media == 1)

{

media = 0;

cout<<"Media unloaded.";

}

else if(media == 0)

{

media = 1;

cout<<"Media loaded.";

}

else if(tray == 0)

cout<<"Tray is not open.";

cout<<"\n\nPress any key to continue.";

getch();

}

};

int main()

{

CDplayer c1;

int s;

while(1)

{

system ("cls");

cout<<"\n\nPlease, Press one of the following buttons: ";

cout<<"\n1. POWER(on/off)";

cout<<"\n2. EJECT(tray open/close)";

cout<<"\n3. PLAY";

cout<<"\n4. STOP";

cout<<"\n5. Load/Unload media to/from the tray(tray will be opened)";

cout<<"\n6. Close CDplayer";

cout<<"\n\nEnter your choice: ";

cin>>s;

switch(s)

{

case 1:

system ("cls");

c1.pushPower();

break;

case 2:

system ("cls");

c1.pushEject();

break;

case 3:

system ("cls");

c1.pushPlay();

break;

case 4:

system ("cls");

c1.pushStop();

break;

case 5:

system ("cls");

c1.loadUnloadMedia();

break;

case 6:

exit(0);

}

getch();

return 0;

}

Output

CD player simulator using c++

Overload - and > operator for class Time

noreply@blogger.com (Maaz Ali) — Thu, 13 Oct 2016 07:33:00 +0000

Problem

Overload - and > operator for class Time, so that this will work:

if (Time2 > Time1)

Time3 = Time2 - Time1;

else

Time3 = Time1 - Time2;

Code

#include "stdafx.h"

#include<conio.h>

#include<iostream>

#include<string.h>

using namespace std;

class time

{

private:

int hr;

int min;

int sec;

public:

time():hr(0),min(0),sec(0)

{ }

void getdata()

{

cout<<"Enter Hours: ";

cin>>hr;

cout<<"Enter Minutes: ";

cin>>min;

cout<<"Enter Seconds: ";

cin>>sec;

}

bool operator > (time t1)

{

if(hr>t1.hr)

return true;

else if(hr == t1.hr)

{

if(min>t1.min)

return true;

else if(min == t1.min)

{

if (sec > t1.sec)

return true;

else if(sec == t1.sec)

{

cout<<"\n\nTimes are EQUAL.";

cout<<"\n\nPress any key to close.";

getch();

exit(0);

}

else

return false;

}

else

return false;

}

else

return false;

}

time operator - (time t1)

{

time temp;

temp.hr = hr - t1.hr;

temp.min = min - t1.min;

temp.sec = sec - t1.sec;

if(temp.min < 0)

temp.min = (temp.min * (-1));

if(temp.sec < 0)

temp.sec = (temp.sec * (-1));

return temp;

}

void display()

{

cout<<hr<<":"<<min<<":"<<sec;

}

};

int main()

{

time time1,time2,time3;

cout<<"Enter 1st Time: \n";

time1.getdata();

cout<<"\n\nEnter 2nd Time: \n";

time2.getdata();

if(time2>time1)

{

time3 = time2 - time1;

cout<<"\n\n2nd Time is greater.";

cout<<"\nTime after subtracting: ";time3.display();

}

else

{

time3 = time1 - time2;

cout<<"\n\n1st Time is greater.";

cout<<"\nTime after subtracting: ";time3.display();

}

cout<<"\n\nPress any key to close.";

getch();

return 0;

}

Output

Class of Employee with static members to count number of employees in the particular department.

noreply@blogger.com (Maaz Ali) — Thu, 13 Oct 2016 07:20:00 +0000

Problem

Create a Class of Employee with static members to count number of employees in the particular department?

Code

#include "stdafx.h"

#include<conio.h>

#include<iostream>

#include<string.h>

using namespace std;

class employee

{

private:

static int cs;

static int phy;

static int maths;

static int eng;

static int mtr;

public:

void comp()

{ cs++; }

void PHY()

{ phy++; }

void MATHS()

{ maths++; }

void ENG()

{ eng++; }

void MTR()

{ mtr++; }

void display()

{

cout<<"\nNo. of employees in cs department ="<<cs;

cout<<"\nNo. of employees in physics department ="<<phy;

cout<<"\nNo. of employees in maths department ="<<maths;

cout<<"\nNo. of employees in engineering department="<<eng;

cout<<"\nNo. of employees in metrology department = "<<mtr;

cout<<"\n\nTotal no. of employees= "<<(cs+phy+maths+eng+mtr);

}

};

int employee::cs=0;

int employee::phy=0;

int employee::maths=0;

int employee::eng=0;

int employee::mtr=0;

int main()

{

employee c[15],phy[15],mth[15],eng[15],mtr[15],e;

int i,n,j;

cout<<"Enter number of employees of cs department: ";

cin>>n;

for(i=0;i<n;i++)

{ c[i].comp(); }

cout<<"Enter number of employees of physics department: ";

cin>>n;

for(i=0;i<n;i++)

{ phy[i].PHY(); }

cout<<"Enter number of employees of maths department: ";

cin>>n;

for(i=0;i<n;i++)

{ c[i].MATHS(); }

cout<<"Enter number of employees of engineering department: ";

cin>>n;

for(i=0;i<n;i++)

{ c[i].ENG(); }

cout<<"Enter number of employees of metrology department: ";

cin>>n;

for(i=0;i<n;i++)

{ c[i].MTR(); }

e.display();

cout<<"\n\nPress any key to close.";

getch();

return 0;

}

Output

calculate students grade and display using assembly language

noreply@blogger.com (Maaz Ali) — Mon, 10 Oct 2016 07:16:00 +0000

Problem

Calculate grade of the students by adding subject marks input by the user. Display the grade to the user.

Code

include irvine32.inc

.data

array1 sdword 10 dup(?)

array2 sdword 10 dup(?)

array3 sdword 10 dup(?)

array4 sdword +100,+100,+100,+100,+100

array5 sdword +100,+100,+2,+2,+100

mul_array sdword 32 dup(0)

mine dword ?

temp dword 0

temp1 dword 0

temp2 dword 0

two dword 2

student1_msg byte " -*-*-*-*-*- Student # 01 -*-*-*-*-*-*-",0

student2_msg byte " -*-*-*-*-*- Student # 02 -*-*-*-*-*-*-",0

la_marks_msg byte " LA = ",0

ms_marks_msg byte " MS = ",0

ds_marks_msg byte " DS = ",0

algo_marks_msg byte " ALGO = ",0

coal_marks_msg byte " COAL = ",0

grade_msg byte "Grade = ",0

grade_f_msg byte "F",0

grade_c_msg byte "C",0

grade_b_msg byte "B",0

grade_a_msg byte "A",0

num1_msg byte "Enter 1st value: ",0

num2_msg byte "Enter 2nd value: ",0

equation_msg byte "Solving the equation: ",0ah,0dh,0

result byte " Result = ",0

even1_msg byte " 1st number is even.",0ah,0dh,0

even2_msg byte " 2nd number is even.",0ah,0dh,0ah,0dh,0

pos_msg byte "1st number is positive.",0ah,0dh,0

pos1_msg byte "2nd number is positive.",0ah,0dh,0

odd_msg byte " one number is odd.",0ah,0dh,0ah,0dh,0

neg_msg byte "number is negative.",0ah,0dh,0

ans_msg byte "Answer = ",0

.code

main proc

comment !

mov esi, offset array1

mov edi, offset array2

mov ebp, offset array3

mov ecx, 10

main_loop:

mov edx, offset num1_msg

call writestring

call readint

mov [esi], eax

mov edx, offset num1_msg

call writestring

call readint

mov [edi], eax

mov eax, [esi]

cmp eax, 0

jge positive_value1

jmp negative_value1

positive_value1:

mov edx, offset pos_msg

call writestring

mov eax, [edi]

cmp eax, 0

jge positive_value2

jmp negative_value1

positive_value2:

call positive

jmp jmp_to_loop

negative_value1:

call negative

jmp_to_loop:

add esi, 4

add edi, 4

add ebp, 4

loop main_loop

mov esi, offset array4

mov edi, offset array5

mov ecx, 5

mov ebp, offset array3

l1:

call crlf

mov eax, [ebp]

;call writeint

mov [esi], eax

add ebp, 4

add esi,4

loop l1

mov ecx, 5

l2:

call crlf

mov eax, [ebp]

;call writeint

mov [edi], eax

add ebp, 4

add edi, 4

loop l2

mov esi, offset array4

mov ecx, 5

l3:

call crlf

mov eax, [esi]

cmp eax, 0

jl negative_to_positive

mov eax, [esi]

cmp eax, 100

ja make_it_100

jmp normal_execution

negative_to_positive:

mov ebx, -1

imul ebx

cmp eax, 100

ja make_it_100

jmp normal_execution

make_it_100:

mov eax, 100

mov [esi], eax

normal_execution:

call writeint

add esi, 4

loop l3

call crlf

mov edi, offset array5

mov ecx, 5

l4:

call crlf

mov eax, [edi]

cmp eax, 0

jl negative_to_positive_1

mov eax, [edi]

cmp eax, 100

ja make_it_100_1

jmp normal_execution_1

negative_to_positive_1:

mov ebx, -1

imul ebx

cmp eax, 100

ja make_it_100

jmp normal_execution_1

make_it_100_1:

mov eax, 100

mov [edi], eax

normal_execution_1:

call writeint

add edi, 4

loop l4

mov edx, offset student1_msg

call writestring

call crlf

mov edx, offset coal_marks_msg

call writestring

mov eax, array4

call writeint

call crlf

mov edx, offset ms_marks_msg

call writestring

mov eax, [array4+4]

call writeint

call crlf

mov edx, offset ds_marks_msg

call writestring

mov eax, [array4+8]

call writeint

call crlf

mov edx, offset algo_marks_msg

call writestring

mov eax, [array4+12]

call writeint

call crlf

mov edx, offset la_marks_msg

call writestring

mov eax, [array4+16]

call writeint

call crlf

mov edx, offset grade_msg

call writestring

mov esi, offset array4

mov ecx, 5

mov eax, 0

l5:

mov ebx, [esi]

add eax, ebx

;call writeint

add esi, 4

loop l5

cmp eax, 399

ja grade_A

cmp eax, 299

ja grade_B

cmp eax, 199

ja grade_C

cmp eax, 0

ja grade_F

jmp grading_L3

grade_A:

mov edx, offset grade_a_msg

call writestring

jmp grading_L3

grade_B:

mov edx, offset grade_b_msg

call writestring

jmp grading_L3

grade_C:

mov edx, offset grade_c_msg

call writestring

jmp grading_L3

grade_F:

mov edx, offset grade_f_msg

call writestring

grading_L3:

call crlf

mov edx, offset student2_msg

call writestring

call crlf

mov edx, offset coal_marks_msg

call writestring

mov eax, array5

call writeint

call crlf

mov edx, offset ms_marks_msg

call writestring

mov eax, [array5+4]

call writeint

call crlf

mov edx, offset ds_marks_msg

call writestring

mov eax, [array5+8]

call writeint

call crlf

mov edx, offset algo_marks_msg

call writestring

mov eax, [array5+12]

call writeint

call crlf

mov edx, offset la_marks_msg

call writestring

mov eax, [array5+16]

call writeint

call crlf

mov edx, offset grade_msg

call writestring

mov edi, offset array5

mov ecx, 5

mov eax, 0

l6:

mov ebx, [edi]

add eax, ebx

add edi,4

;call writeint

loop l6

cmp eax, 399

ja grade_A_1

cmp eax, 299

ja grade_B_1

cmp eax, 199

ja grade_C_1

cmp eax, 0

ja grade_F_1

jmp grading_L4

grade_A_1:

mov edx, offset grade_a_msg

call writestring

jmp grading_L4

grade_B_1:

mov edx, offset grade_b_msg

call writestring

jmp grading_L4

grade_C_1:

mov edx, offset grade_c_msg

call writestring

jmp grading_L4

grade_F_1:

mov edx, offset grade_f_msg

call writestring

grading_L4:

call crlf

;call equation

call crlf

exit

main endp

positive proc

mov edx, offset pos1_msg

call writestring

mov edx, 0

mov eax, [esi]

cdq

idiv two

cmp edx, 0

je even_value1

jmp odd_value1

even_value1:

call crlf

mov edx, offset even1_msg

call writestring

mov edx, 0

mov eax, [edi]

cdq

idiv two

cmp edx, 0

je even_value2

jmp odd_value1

even_value2:

mov edx, offset even2_msg

call writestring

call multiply

call display

jmp jmp_from_even

odd_value1:

mov edx, offset odd_msg

call writestring

call divide

call display

jmp_from_even:

ret

positive endp

negative proc

call crlf

mov edx, offset neg_msg

call writestring

mov edx, 0

mov eax, [esi]

cdq

idiv two

cmp edx, 0

je even_value3

pop edx

jmp odd_value3

even_value3:

mov edx, offset even1_msg

call writestring

mov edx, 0

mov eax, [edi]

cdq

idiv two

cmp edx, 0

je even_value4

jmp odd_value3

even_value4:

mov edx, offset even2_msg

call writestring

call multiply

call display

jmp jmp_from_even

odd_value3:

mov edx, offset odd_msg

call writestring

call divide

call display

jmp_from_even:

ret

negative endp

multiply proc

push ecx

mov edx, offset mul_array

mov ebx, [esi]

mov mine, 0

mov temp, 0

mov ecx, 32

shr ebx,1

jc one_jump

jmp else_jump

one_jump:

mov eax, mine

mov temp, eax

mov eax, [edi]

l1:

cmp temp, 0

je outside_loop

sub temp, 1

shl eax, 1

jmp l1

outside_loop:

mov [edx], eax

add edx, 4

else_jump:

add mine, 1

loop l

mov ecx, lengthof mul_array

mov edx, offset mul_array

mov eax, 0

add_array:

add eax,[edx]

add edx, 4

loop add_array

mov ecx, 32

mov ebx, 0

mov edx, offset mul_array

empty_mul_array:

mov [edx],ebx

add edx, 4

loop empty_mul_array

pop ecx

ret

multiply endp

divide proc

push ecx

mov edx, 0

mov eax, [esi]

cdq

mov ecx, [edi]

idiv ecx

pop ecx

ret

divide endp

display proc

call crlf

mov edx, offset ans_msg

call writestring

call writeint

call crlf

mov [ebp], eax

ret

display endp

equation proc

mov edx, 0

mov eax, [array3+24]

cdq

idiv [array3+12]

mov temp2, eax

mov esi, offset temp2

mov edx, 0

mov eax, array3

cdq

idiv [array3+4]

mov temp1, eax

mov edi, offset temp1

call multiply

mov edx, [array3+16]

add edx, [array3+20]

add eax, edx

mov temp1, eax

mov esi, offset temp1

mov edi, offset [array3+8]

call multiply

add eax, array3[28]

call crlf

mov edx, offset equation_msg

call writestring

call crlf

mov edx, offset result

call writestring

call writeint

ret

equation endp

end main

Output

take input from user and take decision on the basis of these numbers whether to multiply or divide or shift

noreply@blogger.com (Maaz Ali) — Mon, 10 Oct 2016 07:06:00 +0000

Problem:

Taking two inputs from user and taking decision on the basis of these numbers:

If both numbers are positive and even, than multiply those using shift command.
It both numbers are positive but one of them is odd, than divide those using div or idiv.
If one is negative and both numbers are even, than again multiply using shift.
If one number is negative and one of them is odd than divide using idiv or div.

Code

include irvine32.inc

.data

array1 sdword 10 dup(?)
array2 sdword 10 dup(?)
array3 sdword 10 dup(?)

array4 sdword 5 dup(?)
array5 sdword 5 dup(?)

mul_array sdword 32 dup(0)

mine dword ?
temp dword 0
temp1 dword 0
temp2 dword 0

two dword 2

num1_msg byte "Enter 1st value: ",0
num2_msg byte "Enter 2nd value: ",0

equation_msg byte "Solving the equation: ",0ah,0dh,0
result byte " Result = ",0

even1_msg byte " 1st number is even.",0ah,0dh,0
even2_msg byte " 2nd number is even.",0ah,0dh,0ah,0dh,0

pos_msg byte "1st number is positive.",0ah,0dh,0
pos1_msg byte "2nd number is positive.",0ah,0dh,0

odd_msg byte " one number is odd.",0ah,0dh,0ah,0dh,0

neg_msg byte "number is negative.",0ah,0dh,0

ans_msg byte "Answer = ",0

.code
main proc

mov esi, offset array1
mov edi, offset array2
mov ebp, offset array3

mov ecx, 10

main_loop:

mov edx, offset num1_msg
call writestring
call readint

mov [esi], eax

mov edx, offset num1_msg
call writestring
call readint

mov [edi], eax

mov eax, [esi]
cmp eax, 0

jge positive_value1
jmp negative_value1

positive_value1:

mov edx, offset pos_msg
call writestring

mov eax, [edi]
cmp eax, 0

jge positive_value2
jmp negative_value1

positive_value2:
call positive
jmp jmp_to_loop

negative_value1:
call negative

jmp_to_loop:
add esi, 4
add edi, 4
add ebp, 4
loop main_loop

mov esi, offset array4
mov edi, offset array5

mov ecx, 5
mov ebp, offset array3

l1:
call crlf
mov eax, [ebp]
;call writeint
mov [esi], eax
add ebp, 4
add esi,4

loop l1

mov ecx, 5

l2:
call crlf
mov eax, [ebp]
;call writeint
mov [edi], eax
add ebp, 4
add edi, 4
loop l2

mov esi, offset array4
mov ecx, 5

l3:
call crlf
mov eax, [esi]

call writeint
add esi, 4
loop l3

mov edi, offset array5
mov ecx, 5

l4:
call crlf
mov eax, [edi]
call writeint
add edi, 4
loop l4

;call equation

call crlf
call crlf
call crlf
exit
main endp

positive proc

mov edx, offset pos1_msg
call writestring

mov edx, 0
mov eax, [esi]
cdq
idiv two

cmp edx, 0
je even_value1
jmp odd_value1

even_value1:
call crlf
mov edx, offset even1_msg
call writestring

mov edx, 0
mov eax, [edi]
cdq
idiv two

cmp edx, 0
je even_value2
jmp odd_value1

even_value2:
mov edx, offset even2_msg
call writestring
call multiply
call display
jmp jmp_from_even

odd_value1:
mov edx, offset odd_msg
call writestring
call divide
call display

jmp_from_even:

ret
positive endp

negative proc

call crlf
mov edx, offset neg_msg
call writestring

mov edx, 0
mov eax, [esi]
cdq
idiv two

cmp edx, 0
je even_value3
pop edx
jmp odd_value3

even_value3:
mov edx, offset even1_msg
call writestring

mov edx, 0
mov eax, [edi]
cdq
idiv two

cmp edx, 0
je even_value4
jmp odd_value3

even_value4:
mov edx, offset even2_msg
call writestring
call multiply
call display
jmp jmp_from_even

odd_value3:
mov edx, offset odd_msg
call writestring
call divide
call display
jmp_from_even:
ret
negative endp

multiply proc
push ecx

mov edx, offset mul_array

mov ebx, [esi]

mov mine, 0
mov temp, 0
mov ecx, 32

l:
shr ebx,1
jc one_jump

jmp else_jump

one_jump:

mov eax, mine
mov temp, eax

mov eax, [edi]

l1:
cmp temp, 0
je outside_loop
sub temp, 1
shl eax, 1
jmp l1
outside_loop:
mov [edx], eax
add edx, 4
else_jump:
add mine, 1
loop l

mov ecx, lengthof mul_array
mov edx, offset mul_array
mov eax, 0

add_array:

add eax,[edx]
add edx, 4

loop add_array

mov ecx, 32
mov ebx, 0
mov edx, offset mul_array

empty_mul_array:

mov [edx],ebx
add edx, 4

loop empty_mul_array

pop ecx

ret
multiply endp

divide proc
push ecx

mov edx, 0
mov eax, [esi]
cdq
mov ecx, [edi]
idiv ecx

pop ecx
ret
divide endp

display proc

call crlf
mov edx, offset ans_msg
call writestring
call writeint
call crlf
call crlf
mov [ebp], eax

ret
display endp

equation proc

mov edx, 0
mov eax, [array3+24]
cdq
idiv [array3+12]

mov temp2, eax
mov esi, offset temp2

mov edx, 0
mov eax, array3
cdq
idiv [array3+4]

mov temp1, eax

mov edi, offset temp1

call multiply

mov edx, [array3+16]
add edx, [array3+20]

add eax, edx

mov temp1, eax

mov esi, offset temp1
mov edi, offset [array3+8]
call multiply

add eax, array3[28]

call crlf
call crlf

mov edx, offset equation_msg
call writestring
call crlf

mov edx, offset result
call writestring
call writeint

ret
equation endp

end main

Output

Linear Search and Binary Search using C and C++ Language - recursive

noreply@blogger.com (Maaz Ali) — Fri, 07 Oct 2016 12:40:00 +0000

Write programs in C/C++ language to find a specific number from a user input array elements by following linear search and binary search algorithms.

Method – 1: Linear search

Code

#include<conio.h>
#include<iostream>
using namespace std;

int main()
{
//declaring variables and array
int max = 5;
int i,j=0,s;
int a[max];
//Taking input(Numbers) from user.
for(i=0;i<max;i++)
{
cout<<"Enter "<<(i+1)<<"th Number: ";
cin>>a[i];
}
//Asking user to enter the number he/she wants to search
cout<<"\nEnter number to search: ";
cin>>s;
//Searching the number
cout<<"\nAfter Searching:";
for(i=0;i<max;i++)
{
if (a[i]==s)
{
cout<<"\n\tThe number ("<<a[i]<<") is at "<<(i+1)<<"th position.";
j = 1;
}
}

//Display Number and its position (after searching)
if (j == 0)
cout<<"Entered number is not in the list.";

cout<<"\n\nPress any key to exit.";
getch();
return 0;
}

Output

Method – 2: Recursive linear search

Code

#include <iostream>
#include <conio.h>
using namespace std;

int linearSearch( const int array[], int length, int value);

int main()
{
const int arraySize = 100;
int a[arraySize];
int element;

for( int i = 0; i < arraySize; i++)
{
a[i] = 2 * i;
}

element = linearSearch( a, arraySize, 8);

if( element != -1 )
cout << "Found value in element " << element << endl;

else
cout << "Value not found." << endl;

getch();
return 0;
}

int linearSearch( const int array[], int length, int value)
{
if(length==0)
return -1;
else if(array[length-1]==value)
return length-1;
else return linearSearch( array, length-1, value);
}

Output

Method – 3: Binary Search

Code

#include<conio.h>
#include<iostream>
using namespace std;

int main()
{
//declaring variables and array
int max = 10;
int i,j,s,m,location,k;
int a[max];
//Taking input(Numbers) from user.
for(i=0;i<max;i++)
{
cout<<"Enter "<<(i+1)<<"th Number: ";
cin>>a[i];
}
//Asking user to enter the number he/she wants to search
cout<<"\nEnter number to search: ";
cin>>s;
//Searching the number
i = 0;
j = max;
while (i < j)
{
m = ((i+j)/2);
if (s > a[m])
i = m+1;
else
j = m;
}

if (s == a[i])
{ //if searched number is equal to a[i]
cout<<"\n\nEntered number ("<<s<<") is at "<<(i+1)<<"th position.";
}
else
{ //if the searched number is not in the list
cout<<"\n\nEntered number is not in the list.";
}

cout<<"\n\nPress any key to exit.";
getch();
return 0;
}

Output

Find the nth Fibonacci number in C/C++ Language

noreply@blogger.com (Maaz Ali) — Fri, 07 Oct 2016 07:38:00 +0000

There are multiple ways to achieve that. Following are the 3 methods to find nth Fibonacci number.

Algorithm: We know that

Fib (n) = fib (n-1) + fib (n-2)

Our program will stop if one of the following conditions are met:

n = 2 || n = 1

To learn about Fibonacci numbers and sequence (click the link).

Method – 1: iteratively using for loop

Code

#include<iostream>
#include<conio.h>
using namespace std;

int fib(int n);

int main()
{

int n, answer;
cout << "Enter number to find: ";
cin >> n;

cout << "\n\n";

answer = fib(n);

cout << answer << " is the " << n << "th Fibonacci number\n";
getch();
return 0;
}

int fib(int n)
{
int u = 0;
int v = 1;
int i, t;

for(i = 2; i <= n; i++)
{
t = u + v;
u = v;
v = t;
}
return v;
}

Output

Method – 2: Using recursion

Code

#include<iostream>
#include<conio.h>
using namespace std;

int fib(int n);

int main()
{

int n, answer;
cout << "Enter number to find: ";
cin >> n;

cout << "\n\n";

answer = fib(n);

cout << answer << " is the " << n << "th Fibonacci number\n";
getch();
return 0;
}

int fib (int n)
{
cout << "Processing fib(" << n << ")... ";

if (n < 3 )
{
cout << "Return 1!\n";
return (1);
}

else
{
cout << "Call fib(" << n-2 << ") and fib(" << n-1 << ").\n";
return( fib(n-2) + fib(n-1));
}
}

Output

Method – 3: Using recursion

Code

#include<iostream>
#include<conio.h>
using namespace std;

int fib(int n);

int main()
{

int n, answer;
cout << "Enter number to find: ";
cin >> n;

cout << "\n\n";

answer = fib(n);

cout << answer << " is the " << n << "th Fibonacci number\n";
getch();
return 0;
}
int fib(int n)
{
if (n < 2)
return n;
else
return fib(n-1) + fib(n-2);
}

Output

Write a simulator for the assembly language ISA design in C language.

noreply@blogger.com (Maaz Ali) — Wed, 05 Oct 2016 09:46:00 +0000

Code

#include "stdafx.h"

#include <stdio.h>

#include <conio.h>

#include <stdlib.h>

//initializing the memory and registers

void initialization(int a[]){

for(int i=0;i<16;i++)

a[i] = (i+1)*10;

}

//number of instructions

int input_instructions(){

int x;

printf("Enter the number of instructions you want to perform: ");

scanf_s ("%d",&x);

return x;

}

//loading the register from memory

void load(int row, int inst[][3]){

inst[row][0] = 1;

printf("Enter the number of register ranging from 1 to 16: ");

scanf_s("%d",&inst[row][1]);

printf("Enter the number of memory from which you want to load ranging from 1 to 16: ");

scanf_s("%d",&inst[row][2]);

}

//adding the two registers

void add(int row,int inst[][3]){

inst[row][0] = 5;

printf("Enter the destination register number: ");

scanf_s("%d",&inst[row][1]);

printf("Enter the source register number: ");

scanf_s("%d",&inst[row][2]);

}

//storing the value to the memory

void store(int row,int inst[][3]){

inst[row][0] = 3;

printf("Enter the register number: ");

scanf_s("%d",&inst[row][1]);

printf("Enter the memory number: ");

scanf_s("%d",&inst[row][2]);

}

//display the instruction memory

void display_pattern(int no_of_ins,int inst[][3]){

system("cls");

printf("\********************************************************************************");

printf("\n****************************** Instruction Memory ******************************");

printf("\n********************************************************************************");

for(int i=0;i<no_of_ins;i++){

printf("\n\t\t\t\t%d\t%d\t%d\n",inst[i][0],inst[i][1],inst[i][2]);

}

printf("\n\n");

for(int i=0;i<no_of_ins;i++){

if(inst[i][0] == 1){

printf("\t\t\t\tLoadR\tR%d\tM%d\n\n",inst[i][1],inst[i][2]);

}

else if(inst[i][0] == 3){

printf("\t\t\t\tStoreR\tR%d\tM%d\n\n",inst[i][1],inst[i][2]);

}

else if(inst[i][0] == 5){

printf("\t\t\t\tAddR\tR%d\tR%d\n\n",inst[i][1],inst[i][2]);

}

printf("\n\n\n\nPress any key to continue...");

_getch();

}

//display registers and memorys

void display_reg_mem(int reg[],int mem[]){

printf("\n\nRegister: ");

for(int j=0;j<16;j++)

printf("%d ",reg[j]);

printf("\n-----------------------------------------------------------------------------");

printf("\nMemory: ");

for(int j=0;j<16;j++)

printf("%d ",mem[j]);

}

//ISA simulator

void display_cycles(int no_of_inst,int inst[][3],int reg[],int mem[]){

for(int i=0;i<no_of_inst;i++){

system("cls");

printf("********************************************************************************");

printf("\n****************************** ISA Simulator ***********************************");

printf("\n********************************************************************************");

printf("\n\n\n\n\n******************************* Instruction # %d ********************************",(i+1));

printf("\n\n\n\n\nFetch Instruction:\t %d\t%d\t%d\n",inst[i][0],inst[i][1],inst[i][2]);

display_reg_mem(reg,mem);

if(inst[i][0] == 1){

printf("\n\n\n\n\nDecode Instruction:\t LoadR\tRegister%d\tMemory%d\n",inst[i][1],inst[i][2]);

display_reg_mem(reg,mem);

printf("\n\n\n\n\nExecute Instruction:\t Register updated Succesfully\n");

reg[(inst[i][1]-1)] = mem[(inst[i][2]-1)];

display_reg_mem(reg,mem);

}

if(inst[i][0] == 5){

printf("\n\n\n\n\nDecode Instruction:\t AddR\tRegister%d\tRegister%d\n",inst[i][1],inst[i][2]);

display_reg_mem(reg,mem);

printf("\n\n\n\n\nExecute Instruction:\t Register updated Succesfully\n");

reg[(inst[i][1]-1)] = reg[(inst[i][1]-1)] + reg[(inst[i][2]-1)];

display_reg_mem(reg,mem);

}

if(inst[i][0] == 3){

printf("\n\n\n\n\nDecode Instruction:\t StoreR\tRegister%d\tMemory%d\n",inst[i][1],inst[i][2]);

display_reg_mem(reg,mem);

printf("\n\n\n\n\nExecute Instruction:\t Memory updated Succesfully\n");

mem[(inst[i][2]-1)] = reg[(inst[i][1]-1)];

display_reg_mem(reg,mem);

}

printf("\n\n\n\nPress any key to continue...");

_getch();

}

int main(){

system("color 3f");

int choice,count=0,row=0,lines;

int registers[16] = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

int memorys[16];

initialization(memorys);

lines = input_instructions();

int instructions[100][3];

for(int i=0;i<lines;i++){

system("cls");

printf("Select one of the following instruction type from the following:\n");

printf("\t1. Load \n");

printf("\t2. Add \n");

printf("\t3. Store \n");

printf("\t4. Exit \n\n");

printf("Enter your choice: ");

choice = _getche();

switch (choice)

{

case '1': //load

system("cls");

load(row,instructions);

row++;

break;

case '2': //add

system("cls");

add(row,instructions);

row++;

break;

case '3': //store

system("cls");

store(row,instructions);

row++;

break;

case '4': //exit

exit(0);

default:

system("cls");

printf("\n\nInvalid input! Press any key to continue.");

break;

}

system("cls");

display_pattern(lines,instructions);

system("cls");

display_cycles(lines,instructions,registers,memorys);

exit(0);

return 0;

}

Program Output - Screen Shots

Screen 01

In the above screen 01 user must enter the instructions to run the program. For example we want to run our code for 4 instructions, so we will write 4 and hit enter.

The following screen will be displayed.

Screen 02

There are 4 options 1^st option is used to load the register from memory, 2^nd option add the two registers, 3^rd option stores the value from register to the memory, and the 4^th option is used to quit the program. For example, we press 1 and hit enter.

The following screen will be displayed.

Screen 03

User will be asked to enter the register number and the memory number. We will enter 2 for register and 7 for memory.

Screen 04

Same procedure will be followed again i.e. steps for screen 02.

Screen 05

This time we will enter 3 for register and 6 for the memory.

After you have entered the value in the memory hit enter… the program will show you the following screen.

Screen 06

This time press 2 and hit enter.

Screen 07

It will ask you to enter two register numbers one by one. The 1^st register is the destination and the 2^nd register is the source i.e. after the value is entered, the value of the destination will be overwritten with the sum of the two registers in the add instruction. Here we have entered 2 and 3 register numbers where 2^nd register will be overwritten after the execution with the sum of the register no 2 and 3.

Screen 08

After you enter the register number, the above screen will be displayed again, press 3 this time and hit enter.

Screen 09

The above screen will be shown asking the user to enter the register number from which he/she wants to store data and then the program will ask for the memory number. For instance, we enter 2 for register and 6 for memory and press enter.

Screen 10

As, we have entered 4 instructions (2 load, 1 add and 1 store). The above screen will be displayed asking the user to press any key to continue if he/she have viewed the instructions. After we hit enter the following screen will be displayed.

Screen 11

The program will show the user ISA simulator with the 1^st instruction. As, the 1^st instruction was to load the register from the memory i.e. 1 2 7. After the execution the register’s value will be updated and as we can see in our example it has updated the 2^nd register with the value 70 taken from the memory’s 7^th location.

Let’s have a closer look at the above screen 11, we can see there are three steps involved are involved for each cycle (instruction).

1^st step - the fetch instruction

It fetches the instruction from the memory and loads it into the processor.

2^nd step - the decode instruction

After the 1^st step the 2^nd step takes place: It decodes the instruction after it has been fetched, so that the computer can perform its specific operation.

3^rd step - execute instruction

It executes the instruction and updates the values according to the instruction, for instance in this instruction it will update the value of the 2^nd register.

Screen 12

Similarly 2^nd instruction will be displayed. It will update the 3^rd register with the value 60.

Screen 13

The third instruction was to add the 2^nd and 3^rd register and answer have to be stored in the 2^nd register as it was the destination register. Hence, we can see the value of the 2^nd register have been updated to 130 from 70.

Screen 14

The last and 4^th instruction that we entered was to store the result of add instruction from the 3^rd instruction to the memory. In the above example we have stored the value of the 2^nd register in the 6^th memory location.

The user will be asked to press any key to continue and the program will be terminated as there were no more instructions to show.

Load Add and Store Instructions in Assembly Language | Instruction set Architecture

noreply@blogger.com (Maaz Ali) — Fri, 30 Sep 2016 07:42:00 +0000

We already know (from the previous lectures) how an instruction is formed and what the basic structure of an instruction is. Let’s go through few examples of it.

Instruction Set

An instruction set is composed of two parts:

Opcode
Operand (s)

Opcode

Opcode defines the number of operation supported by this instruction format and used for indexing that operation. For example: 0000 is the opcode for LOADI operation.

Operand (s)

Operands can be a register address, memory address, and immediate value.
In our ISA design, the length of instruction is 16 bits. Such that: 4 bits for opcode, and 12 bits for operand(s).

Opcode (4-bits)

Operands (12-bits)

Operations

For this tutorial we will be focusing on the three basic operations, i.e. how to load, add, and store data in assembly language. Following are the few operations along with their opcodes:

OPCODE	Operation
0000	LOADI
0001	LOADM
0010	STOREI
0011	STORER
0100	ADDI
0101	ADDR
0110	SUBI
…	…
…	…
1111	…

In above operations I is for immediate, M is for memory, and R is for register. Let’s look at the registers and memory first before proceeding to the operations.

Registers

There are 16 registers hence 4 bits are required to address them. Registers are named as R0 to R15.

Register Address	Register
0000	R0
0001	R1
0010	R2
0011	R3
0100	R4
0101	R5
0110	R6
…	…
…	…
1111	R15

Memory

There are 28 memory locations where data may reside. Hence 8 bits are required to address them. Memory locations are named as M0 to M255.

Memory Address	Memory
00000000	M0
00000001	M1
00000010	M2
00000011	M3
00000100	M4
00000101	M5
00000110	M6
…	…
…	…
11111111	M255

LOADM Instruction

Definition

It loads an operand value from a memory location to a register.

Bit allocation

LOADM

MEMORY ADDRESS

4 bits for Opcode
4 bits for Register address
8 bits for Memory address

Working

Consider an example:
Assembly Code: LOADM R4 M6
Machine Code: 0001010000000110

LOADM

In the above instruction operand value i.e. 87 from a memory (M6) copied to a register (R4).

STORER Instruction

Define:

It stores an operand value from a register to a Memory location.

Bit allocation:

STORER

MEMORY ADDRESS

4 bits for Opcode
4 bits for Register address
8 bits for Memory address

Working

Consider an example:
Assembly Code: STORER R4 M6
Machine Code: 0011010000000110

STORER

In the above instruction operand value (67) is stored from a register (R4) to a Memory (M6) location.

ADDR Instruction

Define:

It adds register values and keeps the result in the destination register.

Bit allocation:

ADDR

Reg. Address

4 bits for Opcode
4 bits for each Register address

4 bits for Source
4 bits for Destination

Working

Consider an example:
Assembly Code: ADDR R2 R1
Machine Code: 0101 0010 0001 0000

ADDR

Sample Example

LOADM      R1       M25     ; load from memory M25 and store the value in register R1
LOADM      R2      M69      ; load from memory M69 and store the value in register R2
ADDR         R2       R1          ; add values of register R1 and R2
STORER      R2       M2        ; store the value of register R2 to memory M2

Execution of a Assembly Language Instruction

Our CPU will use three stages for executing a single assembly language instruction. These are:

Fetch:

CPU fetches the instruction from Instruction memory (Maintained in a 2D Array).
Decode:

Operation and operands are determined.
Execute:

Operation is performed on the operands.

To see these steps in practical follow the below program written in c language.

Program

A simulator for the discussed ISA design is written in C language. Click on the below link:

Write A Simulator For The Assembly Language ISA Design In C Language.

Debugging Assembly Language | Introduction to

noreply@blogger.com (Maaz Ali) — Tue, 27 Sep 2016 07:12:00 +0000

If you are interested in writing code on command prompt and like writing code in command prompt than this tutorial is for you. Debugging assembly language is easy. Following picture shows an example of it:

What is Debugging?

Debug is a dos program which allows programmers to see execution of the program closely in depth. It is used to:

Change and see memory contents
Enter and execute programs
Hold the program at a specific instruction to check the data values or change them.

Debugging commands Assembly language

Below are the command examples of debugging an assembly language code:

Debug: Enter a program

Q: Quit the program.

R: See register contents/values.

R<register>: see particular register content/value also change it.

Practice Program: See AX register contents and change the value to 2, 34, 543, and GE.

A: this command assembles the program. You can enter assembly instructions into the memory.

A <starting address>: starting address can be an offset of code segment or you can also specify segment register.

Example – your turn: a 100 or a cs: 100 are the same, type a 100 and debug it.

U: a command that un-assembles the program. You can see machine code of your program by using this command. You can also specify the starting and ending addresses of memory section you want to see the code of, U<starting address> (space) <ending address>.

G: G is a go command that allows the program to run until it reaches the breakpoint. At the end register contents are displayed to the programmer.

F: F command is used to fill into the memory. Format of this command is: F<starting address> <ending address> <data>.

E: E command is used to enter any data or change into the memory. For example: E 200 ‘HELLO WORLD’.

D: D command is used to see the register contents. You can also use this command to fill segment providing offset. For example: F 10 10F FF, fills 16bytes with FF.

Assembly Language Instructions With Examples | Introduction

noreply@blogger.com (Maaz Ali) — Tue, 20 Sep 2016 07:58:00 +0000

Instructions

Instructions are statements that execute when we assemble the program. The instructions (written in assembly language) are translated into machine language bytes, as we have already discussed in the lecture 02. These machine language instructions/bytes are then executed by the CPU.

Example instruction: mov ax, bx

An assembly instruction has 4 basic parts:

Label (optional)
Mnemonic (required)
Operand (depends on the instruction)
Comment (optional)

Syntax of assembly language instruction is as follows:

[label:] mnemonic [operands] [;comment]

Note: We will be using Intel IA-32 for this tutorial and onwards.

Labels

There are two types of labels:

Data Labels
Code Labels

Data Label

Data labels are just like variables in any other language. For example, C language instruction: int myVar 10; we know that myVar is a variable name i.e. data label of int type containing value 10. Similarly, in assembly language we use data labels as variables which hold value (s) for us. For example, count is a variable in the following instruction:

count DWORD 100

Data labels are declared in data segment of the code.

Code Label

Code labels are just the names which are used by other instructions to jump from particular position to that position where data label is located. It must end with a semicolon ( : ). Code labels are declared in code segment. For example:

target: mov ax, bx ... jmp target

target is the data label, which is called by the jmp instruction.

Use: code labels are used as targets for jumping while using loops.

Mnemonics

A small word that acts as an identifier for the instruction. The mnemonics are written in code segment. In following examples mov, sub, add, jmp, call, and mul are the mnemonics:

mov Move/assign one value to another (label)
sub Subtract one value from another
add Adds two values
jmp Jump to a specific location
call Call a procedure/module
mul Multiply two values

Operands

Operand is the remaining part of the instruction. There can be 0 to 3 operands. Where each operand is either a memory operand, a register, an input/output port, or a constant expression.

Following are the operand types:

constant expression
constant
memory (data label)
register

Examples

0 operands
stc ; set Carry flag
1 operand
inc myByte ; memory
inc eax ; register
2 operands
sub myByte,25 ; memory, constant
add ebx,ecx ; register, register
add eax,36 * 25 ; register, constant-expression

Comments

You should have used comments in your programs before. As you know there are two types of comments: 1. Single-line and 2. Multi-line comments. In assembly single-line comments begin with ; (semicolon) whereas, multi-line comments begin with a comment directive and symbol/character selected by the programmer itself and also end with the same symbol/character.

Why you should use comments in your programs?

explains the program's purpose
tricky coding techniques
revision information
by whom and when it was written
application-specific explanations