Lets Think About C & C++

History of the C++ language

2011-08-04T17:26:00.000+05:30

History of the C++ language

In this first C++ programming language tutorial we are going to look at the history of the C++ language.

The C programming language was devised in the early 1970s by Dennis M. Ritchie an employee from Bell Labs (AT&T).

Many other programming languages are derived from the C language. Some did well and some did not. The languages Objective-C and C++ for instance are derived from the C language. Both languages add the “object oriented” element to the language C. One of the most recent languages, that used much of the C language, is Java.

The programming language C++ (originally named “C with Classes”) was devised by Bjarne Stroustrup also an employee from Bell Labs (AT&T). Stroustrup started working on C with Classes in 1979.

The idea of creating a new language originated from a wish, to do things, that were not possible with other languages. He had experience with the language Simula and BCPL (Simula is a slow language, but it had some features that were very helpful for large software development projects. BCPL was to low-level). So he chose to enhance the C language with Simula-like features, because the C language was fast and portable. Stroustrup did not only use features from Simula but also borrowed features from the languages Ada, CLU, ALGOL 68 and ML.

In 1983, the name of the language was changed from C with Classes to C++. (The ++ is C language operator). The first commercial release of the C++ language was in October of 1985. At the same time, the first edition of the book “The C++ Programming Language” was released. (Keep in mind that there was no official standard at that time. So the book became an important reference to the language). In 1989, version 2.0 was released of the C++ language.

In 1990, “The Annotated C++ Reference Manual” was published. This work became the basis for the future standard. (Keep in mind that there were additions to the language after 1990). In 1998, a joint ANSI-ISO committee standardized the C++ language.

As you can see it took a long time (almost 20 years) to come to a standardized version of the language. In 2003 the committee published a corrected version of the C++ standard. The last addition was in 2005. (The last addition is not part of the standard, it is a so called “Library Technical Report”. (TR1 for short).

C++ Compilers [GNU and Visual Studio]

2011-08-03T17:12:00.000+05:30

C++ Compilers [GNU and Visual Studio]

A compiler is a program that translates one language (high level) into another language (e.g., assembly language or machine specific language). A compiler translates source code (plain text) into object code (normally in a form suitable for processing by other programs (like a linker)). The most common reason for wanting to translate source code is to create an executable program.
After the compiler translates the source code in object code, the object(‘s) have to be linked into an executable. This is done by a program called a linker (in most cases the compile stage and link stage are done automatically. It is also possible to do these stages separately).
There are many different compilers available. Some compilers are operating system specific other can be used on different platforms. In the following two sections we take a look at two compilers: GNU compiler and Visual Studio.

GNU Compiler:

The GNU Compiler Collection (GCC) includes front ends for C, C++ , Objective-C, etc., as well as libraries for these languages. GCC can be used on many different platforms, but is commonly used on Unix and Linux.
If you want to use GCC on a windows machine you should take a look at Cygwin. It is also possible to install a Linux partition besides a Windows partition (multi boot system). We recommend you use Ubuntu distribution for this.
You can use your favorite text editor to write your programs in. After writing your program you can use the following command to compile the program:
For c programs:

# gcc -o <output name> <your-source.c>

For c++ programs:

# g++ <your-source.cpp> -o <output name>

On some systems, g++ is also installed with the name c++.
It is also possible to compile C++ programs with gcc, however the use of gcc does not add the C++ library.
g++ is a program that calls GCC
and treats .c, .h and .i files as C++ source files instead of C source files
unless -x is used, and automatically specifies linking against the C++ library.

Visual Studio 2005/2008 (Express edition):

Visual studio is the developer studio from Microsoft. It provides a complete set of development tools to create windows programs in many different languages (like visual basic, visual c++, etc.). The complete developer studio is not free. But it is possible to download an express edition of Visual Studio C++ 2005 or 2008. (You have to register).
If you want to use Visual Studio Express for compiling win32 (console) applications you have to install the platform SDK as well. A complete instruction on how to install can be found here. Note: the tutorials on this site are win32 console applications.
After you installed Visual Studio C++ and the platform SDK you can start your project.
To set-up a win32 console application do the following:

File, New, Project
Project types : Visual C++, win32
Templates : Win32 Console Application
Give the project a name and press OK and then click next.
Check Console Application and Empty project by Application settings.
Click finish.

An empty project is now made. To add a new source file do the following:

In the solution explorer select Sources files and right click on it.
Add, New item, Templates: C++ file (.cpp), Name the file and press add.
File, Save all.

The last thing to do is to set some Project properties:

Project, Properties….
Select General
Set Character from “Use Unicode Character Set” to “Use Multi-Byte Character Set”.
Exit with OK.

You are now ready to program your first program. After writing your program you can compile it by pressing F7.

Important:

The examples included in the C and C++ tutorials are all console programs. That means they use text to communicate. All compilers support the compilation of console programs. Check the user’s manual of your compiler for more info on how to compile them. (It is not doable for us to write this down for every compiler).

C++ First Program by Hello World

2011-08-03T13:46:00.000+05:30

C++ First Program, Hello World

// A hello world program in C++

#include<iostream>

using namespace std;

int main()

    cout << "Hello World!";

    return 0;

We start the tutorial series with one of the simplest programs that can be written in the C++ language.

The hello world program is one of the simplest programs, but it already contains the fundamental components that every C++ program has. Don’t be overwhelmed, we will take a look at the code line by line.

Type the code in your favorite editor (always type, don’t use cut/paste. This is better for learning purposes).
Save the program with the name: hello.cpp or C2Hari.CPP.

After saving the source code you can compile it. It should compile without any errors.

Note: Remember, the examples included in the C and C++ tutorials are all console programs. That means they use text to communicate. All compilers support the compilation of console programs. Check the user manual of you compiler for more information on how to compile them.
(It is not doable for us to write this down for every compiler).

// A hello world program in C++

The first line in our program is a comment line. Every line that starts with two slash signs ( // ) are considered comments and will have no effect on the behavior or outcome of the program. (Words between /* and */ will also be considered as comments (old style comments)). Use comments in your programs to explain difficult sections, but don’t overdo it. It is also common practice to start every program with a brief description on what the program will do.

#include<iostream>

Lines beginning with a pound sign (#) are used by the compilers pre-processor. In this case the directive #include tells the pre-processor to include the iostream standard file. This file iostream includes the declarations of the basic standard input/output library in C++. (See it as including extra lines of code that add functionality to your program).

using namespace std;

All the elements of the standard C++ library are declared within what is called a namespace. In this case the namespace with the name std. We put this line in to declare that we will make use of the functionality offered in the namespace std. This line of code is used very frequent in C++ programs that use the standard library. You will see that we will make use of it in most of the source code included in these tutorials.

int main()

int is what is called the return value (in this case of the type integer. Integer is a whole number). What is used for will be explained further down.

Every program must have a main() function. The main function is the point where all C++ programs start their execution. The word main is followed by a pair of round brackets. That is because it is a function declaration (Functions will be explained in more detail in a later tutorial). It is possible to enclose a list of parameters within the round brackets.

{}

The two curly brackets (one in the beginning and one at the end) are used to indicate the beginning and the end of the function main. (Also called the body of a function). Everything contained within these curly brackets is what the function does when it is called and executed. In the coming tutorials you will see that many other statements make use of curly brackets.

cout << “Hello World”;

This line is a C++ statement. A statement is a simple expression that can produce an effect. In this case the statement will print something to our screen.

cout represents the standard output stream in C++.
(There is also a standard input stream that will be explained in another tutorial). In this a sequence of characters (Hello World) will be send to the standard output stream (in most cases this will be your screen).
The words Hello World have to be between ” “, but the ” ” will not be printed on the screen. They indicate that the sentence begins and where it will end.

cout is declared in the iostream standard file within the std namespace. This is the reason why we needed to include that specific file. This is also the reason why we had to declare this specific namespace.

As you can see the statement ends with a semicolon (;). The semicolon is used to mark the end of the statement. The semicolon must be placed behind all statements in C++ programs. So, remember this. One of the common errors is to forget to include a semicolon after a statement.

return 0;

The return statement causes the main function to finish. You may return a return code (in this case a zero) but it depends on how you start your function main( ). We said that main ( ) will return an int (integer). So we have to return something (in this case a zero). A zero normally indicates that everything went ok and a one normally indicates that something has gone wrong. (This is standard practice. After running an UNIX/Linux program there is often a check on the return code).

Indentations

As you can see the cout and the return statement have been indented or moved to the right side. This is done to make the code more readable. In a program as Hello World, it seems a stupid thing to do. But as the programs become more complex, you will see that it makes the code more readable. (Also you will make fewer errors, like forgetting a curly bracket on the end of a function).

C++ Escape code & Constants & Strings Codes

2011-08-02T19:02:00.000+05:30

C++ constants, escape codes and strings

1.Escape Codes :

Escape codes are used to represent characters that are difficult to express otherwise in the source code. For instance a tab (\t). Escape codes all start with a backslash (\).

\n	Newline
\r	carriage return
\t	Tab
\v	vertical tab
\b	Backspace
\f	form feed (page feed)
\a	alert (beep)
\’	single quote (‘)
\”	double quote (“)
\?	question mark (?)
\\	backslash (\)

Escape codes can be used like so:

        "jane\n"

        "jane\t doe"

        etc.

Escape codes can also be used to express octal (base-8) or hexadecimal (base-16) numbers. An octal number can be used like this: \10 (backslash followed by a number.)

A hexadecimal number can be used like this: \xF0 (a backslash followed by an x and the number.)

It is also possible to define an escape code.

For example:

        #include<iostream>

        using namespace std;

        #define PI 3.14159     // note (1)

        #define NEW_LINE '\n';         // note (2)

        int main ()

               float my_var=5.0;

               my_var = my_var * 2;

               cout << my_var;

               cout << NEW_LINE;

               return 0;

Note: (1) The #define directive is not a C++ statement but a directive for the pre-processor. It assumes the entire line is the directive and does not require a semicolon.

(2) If a semicolon character ( ; ) is appended at the end of the define statement, the pre-processor will replace all NEW_LINE for \n.

2.Constants:

The difference between variables and constants is that variables can change their value at any time but constants can never change their value. (The constants value is locked for the duration of the program.)
Constants can be very useful, PI for instance is a good example to declare as a constant.

Defining a constants:

With the pre-processor directive “define” it is possible to define your own constants.

The format of the pre-processor directive #define is:
#define identifier value

Take a look at an example:

        #define PI 3.14

        #define A 5

Consent Prefix:

The const prefix can be used to declare a constant of a specific type.
(The same way as you would declare any other variable.)

Take a look at the example:

        const int x = 50;

        const char newline = '\n';

        const y = 50;

Note: If no data type is presented (last example) the compiler will assume a data type!
In this case it will assume that the data type is an integer (int).

3.Strings:

Introduction to Strings:

In this paragraph we take a brief look at strings. In a later tutorial we will take a closer look at strings.

Variables that can store more then a single character are known as strings.

The C++ Standard Template Library (STL) contains a string class that has to be included before you can make use of strings. Also you need to have access to the standard namespace.

Take a look at an example:

  #include<iostream>

  #include<string> //add for std::string

  using namespace std;

  int main()

               string mystring = "This is a string";

               cout << mystring;

               return 0;

Note: Instead of “string mystring” you could also say “std::string mystring” if you don’t want to include using namespace std. Instead of mystring = “This is a string”; it is also possible to make use of parentheses: mystring = (“This is a string”);

Strings can also perform all the other basic operations that fundamental data types can. For instance, being declared without an initial value and being assigned values during execution.

Take a look at the following example:

        #include<iostream>

        #include<string>

        using namespace std;

        int main ()

               string mystring;

               mystring = "First value";

               cout << mystring << endl;

               mystring = "Second value";

               cout << mystring << endl;

               return 0;

That’s all for this tutorial. In the next tutorial we will look at operators.

C-Language Aptitude Questions & Answers

2011-08-02T11:06:00.000+05:30

C Language Aptitude FAQ's

1. void main()

{

int const * p=5;

printf("%d",++(*p));

}

Answer:

Compiler error: Cannot modify a constant value.

Explanation:

p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2. main()

{

char s[ ]="man";

int i;

for(i=0;s[ i ];i++)

printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);

}

Answer:

mmmm

aaaa

nnnn

Explanation:

s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3. main()

{

float me = 1.1;

double you = 1.1;

if(me==you)

printf("I love U");

else

printf("I hate U");

}

Answer:

I hate U

Explanation:

For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.

Rule of Thumb:

Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

4. main()

{

static int var = 5;

printf("%d ",var--);

if(var)

main();

}

Answer:

5 4 3 2 1

Explanation:

When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5. main()

{

int c[ ]={2.8,3.4,4,6.7,5};

int j,*p=c,*q=c;

for(j=0;j<5;j++) {

printf(" %d ",*c);

++q; }

for(j=0;j<5;j++){

printf(" %d ",*p);

++p; }

}

Answer:

2 2 2 2 2 2 3 4 6 5

Explanation:

Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

6. main()

{

extern int i;

i=20;

printf("%d",i);

}

Answer:

Linker Error : Undefined symbol '_i'

Explanation:

extern storage class in the following declaration,

extern int i;

specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

7. main()

{

int i=-1,j=-1,k=0,l=2,m;

m=i++&&j++&&k++||l++;

printf("%d %d %d %d %d",i,j,k,l,m);

}

Answer:

0 0 1 3 1

Explanation :

Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

8. main()

{

char *p;

printf("%d %d ",sizeof(*p),sizeof(p));

}

Answer:

1 2

Explanation:

The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

9. main()

{

int i=3;

switch(i)

{

default:printf("zero");

case 1: printf("one");

break;

case 2:printf("two");

break;

case 3: printf("three");

break;

}

Answer :

three

Explanation :

The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

10. main()

{

printf("%x",-1<<4);

}

Answer:

fff0

Explanation :

-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

11. main()

{

char string[]="Hello World";

display(string);

}

void display(char *string)

{

printf("%s",string);

}

Answer:

Compiler Error : Type mismatch in redeclaration of function display

Explanation :

In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

12. main()

{

int c=- -2;

printf("c=%d",c);

}

Answer:

c=2;

Explanation:

Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.

Note:

However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

13. #define int char

main()

{

int i=65;

printf("sizeof(i)=%d",sizeof(i));

}

Answer:

sizeof(i)=1

Explanation:

Since the #define replaces the string int by the macro char

14. main()

{

int i=10;

i=!i>14;

Printf ("i=%d",i);

}

Answer:

i=0

Explanation:

In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

15. #include<stdio.h>

main()

{

char s[]={'a','b','c','\n','c','\0'};

char *p,*str,*str1;

p=&s[3];

str=p;

str1=s;

printf("%d",++*p + ++*str1-32);

}

Answer:

Explanation:

p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.

Now performing (11 + 98 – 32), we get 77("M");

So we get the output 77 :: "M" (Ascii is 77).

16. #include<stdio.h>

main()

{

int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };

int *p,*q;

p=&a[2][2][2];

*q=***a;

printf("%d----%d",*p,*q);

}

Answer:

SomeGarbageValue---1

Explanation:

p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

17. #include<stdio.h>

main()

{

struct xx

{

int x=3;

char name[]="hello";

};

struct xx *s;

printf("%d",s->x);

printf("%s",s->name);

}

Answer:

Compiler Error

Explanation:

You should not initialize variables in declaration

18. #include<stdio.h>

main()

{

struct xx

{

int x;

struct yy

{

char s;

struct xx *p;

};

struct yy *q;

};

}

Answer:

Compiler Error

Explanation:

The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

19. main()

{

printf("\nab");

printf("\bsi");

printf("\rha");

}

Answer:

hai

Explanation:

\n - newline

\b - backspace

\r - linefeed

20. main()

{

int i=5;

printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);

}

Answer:

45545

Explanation:

The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

21. #define square(x) x*x

main()

{

int i;

i = 64/square(4);

printf("%d",i);

}

Answer:

Explanation:

the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

22. main()

{

char *p="hai friends",*p1;

p1=p;

while(*p!='\0') ++*p++;

printf("%s %s",p,p1);

}

Answer:

ibj!gsjfoet

Explanation:

++*p++ will be parse in the given order

Ø *p that is value at the location currently pointed by p will be taken

Ø ++*p the retrieved value will be incremented

Ø when ; is encountered the location will be incremented that is p++ will be executed

Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.

23. #include <stdio.h>

#define a 10

main()

{

#define a 50

printf("%d",a);

}

Answer:

Explanation:

The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24. #define clrscr() 100

main()

{

clrscr();

printf("%d\n",clrscr());

}

Answer:

100

Explanation:

Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :

main()

{

100;

printf("%d\n",100);

}

Note:

100; is an executable statement but with no action. So it doesn't give any problem

25. main()

{

printf("%p",main);

}

Answer:

Some address will be printed.

Explanation:

Function names are just addresses (just like array names are addresses).

main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

27) main()

{

clrscr();

}

clrscr();

Answer:

No output/error

Explanation:

The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

28) enum colors {BLACK,BLUE,GREEN}

main()

{

printf("%d..%d..%d",BLACK,BLUE,GREEN);

return(1);

}

Answer:

0..1..2

Explanation:

enum assigns numbers starting from 0, if not explicitly defined.

29) void main()

{

char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}

Answer:

4..2

Explanation:

the second pointer is of char type and not a far pointer

30) main()

{

int i=400,j=300;

printf("%d..%d");

}

Answer:

400..300

Explanation:

printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.

31) main()

{

char *p;

p="Hello";

printf("%c\n",*&*p);

}

Answer:

Explanation:

* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

32) main()

{

int i=1;

while (i<=5)

{

printf("%d",i);

if (i>2)

goto here;

i++;

}

fun()

{

here:

printf("PP");

}

Answer:

Compiler error: Undefined label 'here' in function main

Explanation:

Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

33) main()

{

static char names[5][20]={"pascal","ada","cobol","fortran","perl"};

int i;

char *t;

t=names[3];

names[3]=names[4];

names[4]=t;

for (i=0;i<=4;i++)

printf("%s",names[i]);

}

Answer:

Compiler error: Lvalue required in function main

Explanation:

Array names are pointer constants. So it cannot be modified.

34) void main()

{

int i=5;

printf("%d",i++ + ++i);

}

Answer:

Output Cannot be predicted exactly.

Explanation:

Side effects are involved in the evaluation of i

35) void main()

{

int i=5;

printf("%d",i+++++i);

}

Answer:

Compiler Error

Explanation:

The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

36) #include<stdio.h>

main()

{

int i=1,j=2;

switch(i)

{

case 1: printf("GOOD");

break;

case j: printf("BAD");

break;

}

Answer:

Compiler Error: Constant expression required in function main.

Explanation:

The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).

Note:

Enumerated types can be used in case statements.

37) main()

{

int i;

printf("%d",scanf("%d",&i)); // value 10 is given as input here

}

Answer:

Explanation:

Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

38) #define f(g,g2) g##g2

main()

{

int var12=100;

printf("%d",f(var,12));

}

Answer:

100

39) main()

{

int i=0;

for(;i++;printf("%d",i)) ;

printf("%d",i);

}

Answer:

Explanation:

before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

40) #include<stdio.h>

main()

{

char s[]={'a','b','c','\n','c','\0'};

char *p,*str,*str1;

p=&s[3];

str=p;

str1=s;

printf("%d",++*p + ++*str1-32);

}

Answer:

Explanation:

p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.

i.e. (11+98-32)=77("M");

41) #include<stdio.h>

main()

{

struct xx

{

int x=3;

char name[]="hello";

};

struct xx *s=malloc(sizeof(struct xx));

printf("%d",s->x);

printf("%s",s->name);

}

Answer:

Compiler Error

Explanation:

Initialization should not be done for structure members inside the structure declaration

42) #include<stdio.h>

main()

{

struct xx

{

int x;

struct yy

{

char s;

struct xx *p;

};

struct yy *q;

};

}

Answer:

Compiler Error

Explanation:

in the end of nested structure yy a member have to be declared.

43) main()

{

extern int i;

i=20;

printf("%d",sizeof(i));

}

Answer:

Linker error: undefined symbol '_i'.

Explanation:

extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

44) main()

{

printf("%d", out);

}

int out=100;

Answer:

Compiler error: undefined symbol out in function main.

Explanation:

The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

45) main()

{

extern out;

printf("%d", out);

}

int out=100;

Answer:

100

Explanation:

This is the correct way of writing the previous program.

46) main()

{

show();

}

void show()

{

printf("I'm the greatest");

}

Answer:

Compier error: Type mismatch in redeclaration of show.

Explanation:

When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.

The solutions are as follows:

1. declare void show() in main() .

2. define show() before main().

3. declare extern void show() before the use of show().

47) main( )

{

int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};

printf(“%u %u %u %d \n”,a,*a,**a,***a);

printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);

}

Answer:

100, 100, 100, 2

114, 104, 102, 3

Explanation:

The given array is a 3-D one. It can also be viewed as a 1-D array.

100 102 104 106 108 110 112 114 116 118 120 122

thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a gives the value. Hence, the first line of the output.

for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.

48) main( )

{

int a[ ] = {10,20,30,40,50},j,*p;

for(j=0; j<5; j++)

{

printf(“%d” ,*a);

a++;

}

p = a;

for(j=0; j<5; j++)

{

printf(“%d ” ,*p);

p++;

}

Answer:

Compiler error: lvalue required.

Explanation:

Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

49) main( )

{

static int a[ ] = {0,1,2,3,4};

int *p[ ] = {a,a+1,a+2,a+3,a+4};

int **ptr = p;

ptr++;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

*ptr++;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

*++ptr;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

++*ptr;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

}

Answer:

111

222

333

344

Explanation:

Let us consider the array and the two pointers with some address

100 102 104 106 108

100

102

104

106

108

1000 1002 1004 1006 1008

ptr

1000

2000

After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.

After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.

After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.

After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.

50) main( )

{

char *q;

int j;

for (j=0; j<3; j++) scanf(“%s” ,(q+j));

for (j=0; j<3; j++) printf(“%c” ,*(q+j));

for (j=0; j<3; j++) printf(“%s” ,(q+j));

}

Explanation:

Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as

M

O

U

When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.

M

T

R

A

C

K

\0

The third input starts filling from the location 102

M

T

V

I

R

T

U

A

L

This is the final value stored .

The first printf prints the values at the position q, q+1 and q+2 = M T V

The second printf prints three strings starting from locations q, q+1, q+2

i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.

51) main( )

{

void *vp;

char ch = ‘g’, *cp = “goofy”;

int j = 20;

vp = &ch;

printf(“%c”, *(char *)vp);

vp = &j;

printf(“%d”,*(int *)vp);

vp = cp;

printf(“%s”,(char *)vp + 3);

}

Answer:

g20fy

Explanation:

Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4^th value hence the output is ‘fy’.

52) main ( )

{

static char *s[ ] = {“black”, “white”, “yellow”, “violet”};

char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;

p = ptr;

**++p;

printf(“%s”,*--*++p + 3);

}

Answer:

Explanation:

In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

53) main()

{

int i, n;

char *x = “girl”;

n = strlen(x);

*x = x[n];

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

{

printf(“%s\n”,x);

x++;

}

Answer:

(blank space)

irl

Explanation:

Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.

54) int i,j;

for(i=0;i<=10;i++)

{

j+=5;

assert(i<5);

}

Answer:

Runtime error: Abnormal program termination.

assert failed (i<5), <file name>,<line number>

Explanation:

asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,

#undef NDEBUG

and this will disable all the assertions from the source code. Assertion

is a good debugging tool to make use of.

55) main()

{

int i=-1;

+i;

printf("i = %d, +i = %d \n",i,+i);

}

Answer:

i = -1, +i = -1

Explanation:

Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

56) What are the files which are automatically opened when a C file is executed?

Answer:

stdin, stdout, stderr (standard input,standard output,standard error).

57) what will be the position of the file marker?

a: fseek(ptr,0,SEEK_SET);

b: fseek(ptr,0,SEEK_CUR);

Answer :

a: The SEEK_SET sets the file position marker to the starting of the file.

b: The SEEK_CUR sets the file position marker to the current position

of the file.

58) main()

{

char name[10],s[12];

scanf(" \"%[^\"]\"",s);

}

How scanf will execute?

Answer:

First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it reads all character upto another quotation mark.

59) What is the problem with the following code segment?

while ((fgets(receiving array,50,file_ptr)) != EOF)

;

Answer & Explanation:

fgets returns a pointer. So the correct end of file check is checking for != NULL.

60) main()

{

main();

}

Answer:

Runtime error : Stack overflow.

Explanation:

main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.

61) main()

{

char *cptr,c;

void *vptr,v;

c=10; v=0;

cptr=&c; vptr=&v;

printf("%c%v",c,v);

}

Answer:

Compiler error (at line number 4): size of v is Unknown.

Explanation:

You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

62) main()

{

char *str1="abcd";

char str2[]="abcd";

printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));

}

Answer:

2 5 5

Explanation:

In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

63) main()

{

char not;

not=!2;

printf("%d",not);

}

Answer:

Explanation:

! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

64) #define FALSE -1

#define TRUE 1

#define NULL 0

main() {

if(NULL)

puts("NULL");

else if(FALSE)

puts("TRUE");

else

puts("FALSE");

}

Answer:

TRUE

Explanation:

The input program to the compiler after processing by the preprocessor is,

main(){

if(0)

puts("NULL");

else if(-1)

puts("TRUE");

else

puts("FALSE");

}

Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

65) main()

{

int k=1;

printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");

}

Answer:

1==1 is TRUE

Explanation:

When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

66) main()

{

int y;

scanf("%d",&y); // input given is 2000

if( (y%4==0 && y%100 != 0) || y%100 == 0 )

printf("%d is a leap year");

else

printf("%d is not a leap year");

}

Answer:

2000 is a leap year

Explanation:

An ordinary program to check if leap year or not.

67) #define max 5

#define int arr1[max]

main()

{

typedef char arr2[max];

arr1 list={0,1,2,3,4};

arr2 name="name";

printf("%d %s",list[0],name);

}

Answer:

Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation:

arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.

Rule of Thumb:

#defines are used for textual replacement whereas typedefs are used for declaring new types.

68) int i=10;

main()

{

extern int i;

{

int i=20;

{

const volatile unsigned i=30;

printf("%d",i);

}

printf("%d",i);

}

printf("%d",i);

}

Answer:

30,20,10

Explanation:

'{' introduces new block and thus new scope. In the innermost block i is declared as,

const volatile unsigned

which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

69) main()

{

int *j;

{

int i=10;

j=&i;

}

printf("%d",*j);

}

Answer:

Explanation:

The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

70) main()

{

int i=-1;

-i;

printf("i = %d, -i = %d \n",i,-i);

}

Answer:

i = -1, -i = 1

Explanation:

-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

71) #include<stdio.h>

main()

{

const int i=4;

float j;

j = ++i;

printf("%d %f", i,++j);

}

Answer:

Compiler error

Explanation:

i is a constant. you cannot change the value of constant

72) #include<stdio.h>

main()

{

int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };

int *p,*q;

p=&a[2][2][2];

*q=***a;

printf("%d..%d",*p,*q);

}

Answer:

garbagevalue..1

Explanation:

p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.

73) #include<stdio.h>

main()

{

char j[]= "hello";

printf("%s %d",j,i);

}

Answer:

hello 5

Explanation:

if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.

74) main()

{

int i=5,j=6,z;

printf("%d",i+++j);

}

Answer:

Explanation:

the expression i+++j is treated as (i++ + j)

76) struct aaa{

struct aaa *prev;

int i;

struct aaa *next;

};

main()

{

struct aaa abc,def,ghi,jkl;

int x=100;

abc.i=0;abc.prev=&jkl;

abc.next=&def;

def.i=1;def.prev=&abc;def.next=&ghi;

ghi.i=2;ghi.prev=&def;

ghi.next=&jkl;

jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;

x=abc.next->next->prev->next->i;

printf("%d",x);

}

Answer:

Explanation:

above all statements form a double circular linked list;

abc.next->next->prev->next->i

this one points to "ghi" node the value of at particular node is 2.

77) struct point

{

int x;

int y;

};

struct point origin,*pp;

main()

{

pp=&origin;

printf("origin is(%d%d)\n",(*pp).x,(*pp).y);

printf("origin is (%d%d)\n",pp->x,pp->y);

}

Answer:

origin is(0,0)

Explanation:

pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.

Note:

Since structure point is globally declared x & y are initialized as zeroes

78) main()

{

int i=_l_abc(10);

printf("%d\n",--i);

}

int _l_abc(int i)

{

return(i++);

}

Answer:

Explanation:

return(i++) it will first return i and then increments. i.e. 10 will be returned.

79) main()

{

char *p;

int *q;

long *r;

p=q=r=0;

p++;

q++;

r++;

printf("%p...%p...%p",p,q,r);

}

Answer:

0001...0002...0004

Explanation:

++ operator when applied to pointers increments address according to their corresponding data-types.

80) main()

{

char c=' ',x,convert(z);

getc(c);

if((c>='a') && (c<='z'))

x=convert(c);

printf("%c",x);

}

convert(z)

{

return z-32;

}

Answer:

Compiler error

Explanation:

declaration of convert and format of getc() are wrong.

81) main(int argc, char **argv)

{

printf("enter the character");

getchar();

sum(argv[1],argv[2]);

}

sum(num1,num2)

int num1,num2;

{

return num1+num2;

}

Answer:

Compiler error.

Explanation:

argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.

82) # include <stdio.h>

int one_d[]={1,2,3};

main()

{

int *ptr;

ptr=one_d;

ptr+=3;

printf("%d",*ptr);

}

Answer:

garbage value

Explanation:

ptr pointer is pointing to out of the array range of one_d.

83) # include<stdio.h>

aaa() {

printf("hi");

}

bbb(){

printf("hello");

}

ccc(){

printf("bye");

}

main()

{

int (*ptr[3])();

ptr[0]=aaa;

ptr[1]=bbb;

ptr[2]=ccc;

ptr[2]();

}

Answer:

bye

Explanation:

ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

85) #include<stdio.h>

main()

{

FILE *ptr;

char i;

ptr=fopen("zzz.c","r");

while((i=fgetch(ptr))!=EOF)

printf("%c",i);

}

Answer:

contents of zzz.c followed by an infinite loop

Explanation:

The condition is checked against EOF, it should be checked against NULL.

86) main()

{

int i =0;j=0;

if(i && j++)

printf("%d..%d",i++,j);

printf("%d..%d,i,j);

}

Answer:

0..0

Explanation:

The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain unchanged and get printed.

87) main()

{

int i;

i = abc();

printf("%d",i);

}

abc()

{

_AX = 1000;

}

Answer:

1000

Explanation:

Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.

88) int i;

main(){

int t;

for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))

printf("%d--",t--);

}

// If the inputs are 0,1,2,3 find the o/p

Answer:

4--0

3--1

2--2

Explanation:

Let us assume some x= scanf("%d",&i)-t the values during execution

will be,

t i x

4 0 -4

3 1 -2

2 2 0

89) main(){

int a= 0;int b = 20;char x =1;char y =10;

if(a,b,x,y)

printf("hello");

}

Answer:

hello

Explanation:

The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.

90) main(){

unsigned int i;

for(i=1;i>-2;i--)

printf("c aptitude");

}

Explanation:

i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.

91) In the following pgm add a stmt in the function fun such that the address of

'a' gets stored in 'j'.

main(){

int * j;

void fun(int **);

fun(&j);

}

void fun(int **k) {

int a =0;

/* add a stmt here*/

}

Answer:

*k = &a

Explanation:

The argument of the function is a pointer to a pointer.

92) What are the following notations of defining functions known as?

i. int abc(int a,float b)

{

/* some code */

}

ii. int abc(a,b)

int a; float b;

{

/* some code*/

}

Answer:

i. ANSI C notation

ii. Kernighan & Ritche notation

93) main()

{

char *p;

p="%d\n";

p++;

printf(p-2,300);

}

Answer:

300

Explanation:

The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.

94) main(){

char a[100];

a[0]='a';a[1]]='b';a[2]='c';a[4]='d';

abc(a);

}

abc(char a[]){

a++;

printf("%c",*a);

a++;

printf("%c",*a);

}

Explanation:

The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so bc will be printed.

95) func(a,b)

int a,b;

{

return( a= (a==b) );

}

main()

{

int process(),func();

printf("The value of process is %d !\n ",process(func,3,6));

}

process(pf,val1,val2)

int (*pf) ();

int val1,val2;

{

return((*pf) (val1,val2));

}

Answer:

The value if process is 0 !

Explanation:

The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function 'process'.

96) void main()

{

static int i=5;

if(--i){

main();

printf("%d ",i);

}

Answer:

0 0 0 0

Explanation:

The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

97) void main()

{

int k=ret(sizeof(float));

printf("\n here value is %d",++k);

}

int ret(int ret)

{

ret += 2.5;

return(ret);

}

Answer:

Here value is 7

Explanation:

The int ret(int ret), ie., the function name and the argument name can be the same.

Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.

98) void main()

{

char a[]="12345\0";

int i=strlen(a);

printf("here in 3 %d\n",++i);

}

Answer:

here in 3 6

Explanation:

The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.

99) void main()

{

unsigned giveit=-1;

int gotit;

printf("%u ",++giveit);

printf("%u \n",gotit=--giveit);

}

Answer:

0 65535

Explanation:

100) void main()

{

int i;

char a[]="\0";

if(printf("%s\n",a))

printf("Ok here \n");

else

printf("Forget it\n");

}

Answer:

Ok here

Explanation:

Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.

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2011-07-25T12:50:00.001+05:30

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2011-07-25T12:48:00.000+05:30

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