Test your C/C++ Programming Skills

Test your C++ Programming skills Part 1

2011-11-27T20:10:00.000-08:00

Q1. The class that in C++ for file input is?

Ans:

ifstream

Q2. How would you define a class with a static data member and initialise it correctly?

Ans:

class bad { static I;}
int bad::I=100;

Q3. What is the error in the code snippet below?
int main()
{
int *p;
int q;
p = new int;
delete q;
return 0;
}

Ans:

"p" cannot be initialized to "new int" since "p" is an integer pointer.

Q4.
class Foo {

   char *buf;
public:
   Foo (char const * b = "default") {
      buf = new char[srd::strlen(b)+1];
      std::strcpy(buf, b);
   }

   ~Foo() {
     delete[] buf;
   }
};

Foo Func(Foo f) {
   return f;
}

Referring to the code above, when the "func" is called, the program may crash or have any unexpected behaviour. What is the reason of this problem?

Ans:

Try yourself

Q5. Find the o/p:
#include < cstudio >

cclass A {
public:
   A() { std::putchar('A); }
   ~A() { std::putchar('~); std::putchar('A); }
}

class B {
public:
   B() { std::putchar('B); }
   ~B() {std::putchar('~); std::putchar('B); }
}

A a;
static B b;
int main() {
   static B b; return 0;
}

Ans:

Try yourself...

Q6.
class A {
public:
   A() { std::memset(buf, 0, sizeof(buf)); }

protected:
   char buf[256];
};

class B {
public:
   B(const A& arg) : a(arg) {} //Ctr1
   B(const A& arg) { a = arg;} //Ctr2

protected:
   A a;
};

Consider the two constructors given for B in the sample. Only one of the constructors can be used because they have the same signature, which one?

Ans:

Try yourself...

Q7. Choose the correct option:

Mutable keyword can be applied to:
a) non-static data members of a class
b) non-const data members of a class
c) Both a and b
d) None of above

Ans:

Try yourself...

Q8. Choose the correct option:
Objects have
a) Behaviour
b) State
c) both a and b
d) None

Ans:

You tell me...

Q9. The vtable entry for a pure virtual function in C++ is:
a) Zero
b) NULL
c) One
d) No entry in vtable

Ans:

Try yourself

Q10. Find the o/p of this:
main() {

   int i;
   float *pf;

   pf = (float *) &i;
   *pf = 100.00;
}

Ans:

Runtime error

Q11. Can we declare a static function as virtual?

Ans:

No.
The virtual function mechanism is used on the specific object that determines which virtual function to call. Since the static functions are not any way related to objects, they can not be declared as virtual.

Q12. Is it compulsory to override a virtual function?

Ans:

No.
If we do not override the virtual function in the derived class then the VTABLE of the derived class would contain the address of the base class virtual function.

Q13.Can we get the typeid of void pointers?

Ans:

No.
At runtime, type identification doesn't work with void pointers, because void * truly means no type information at all.

Test your C Programming skills Part 4

2011-11-25T20:02:00.000-08:00

Q1. How scanf will execute?
main(){

char name[10],s[12];
scanf(" \"%[^\"]\"",s);

}

Ans:

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.

Q2. What will be the position of the file marker?

a) fseek(ptr,0,SEEK_SET);

b) fseek(ptr,0,SEEK_CUR);

Ans:

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.

Q3. What is the problem with the following code segment?

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

Ans:

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

NULL.

Q4. Find the o/p:
main(){

   char *cptr,c;
   void *vptr,v;

   c=10; v=0;
   cptr=&c; vptr=&v;

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

}

Ans:

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.

Q5. Find the o/p:
#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);

}
Ans:

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

Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for
declaring new types.

Q6. Find the o/p:
int i=10;

main() {

   extern int i; {

   int i=20;{
     int i=30;
     printf("%d",i);

   }

   printf("%d",i);

   }

   printf("%d",i);

}
Ans:

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.

Q7. Find the o/p:
main(){

   int *j;{
     int i=10;
     j=&i;
   }

   printf("%d",*j);

}

Ans:

10

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.

Q8. Find the o/p:
# include < stdio.h >
main() {

const int i=4;

float j;

j = ++i;

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

}
Ans:

Compiler error

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

Q9. Find the o/p:
main() {

int i=_l_abc(10);

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

}

int _l_abc(int i) {

return(i++);

}

Ans:

9

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

Q10. Find the o/p of this:
main(int argc, char **argv) {

   printf("enter the character");
   getchar();

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

}

sum(num1,num2)int num1,num2; {

return num1+num2;

}
Ans:

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

Q11. Find the o/p:
int one_d[]={1,2,3};

main(){

int *ptr;

   ptr=one_d;
   ptr+=3;

   printf("%d",*ptr);
}

Ans:

garbage value

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

Q12. Find the o/p:
main(){

FILE *ptr;
char i;

   ptr=fopen("zzz.c","r");
   while((i=fgetch(ptr))!=EOF)
      printf("%c",i);

}

Ans:

contents of zzz.c followed by an infinite loop

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

Q13.Find the o/p:
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

Ans:

4--0

3--1

2--2

Test your C Programming skills Part 3

2011-11-25T19:37:00.000-08:00

Q1. Find the o/p:
main() {

   char *p;

   p="Hello";

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

}
Ans:

Q2. Find the o/p:
main() {

   int i=1;

   while (i<=5){      printf("%d",i);      if (i>2)

     i++;

   }

}

fun() {

   here:

   printf("PP");

}
Ans:

Compiler error: Undefined label 'here' in function main

Q3. Find the o/p:
void main() {

int i=5;

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

}

Ans:

     Output Cannot be predicted exactly.

     Explanation:
     Side effects are involved in the evaluation of i.

Q4. Find the o/p:
main() {

   int i=1,j=2;

   switch(i) {

   case 1: printf("GOOD");

   break;

   case j: printf("BAD");

   break;

   }

}
Ans:

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.

Q5. Find the o/p:
main() {

int i;

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

}

Ans:

1

Explanation:
Scanf returns number of items successfully read.Here 10 is given as input
which should have been scanned successfully. So number of items read is 1.

Q6. Find the o/p:
#define f(g,g2) g##g2

main() {

int var12=100;

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

}
Ans:

100

Q7. Find the o/p:
main() {

int i=0;

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

}
Ans:

Q8. Find the o/p:
# include < stdio.h >
main() {

printf("%d", out);

}

int out=100;

Ans:

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.

Q9. Find the o/p:
#include <stdio.h>
main() {

extern out;

printf("%d", out);

}

int out=100;

Ans:

100

Explanation:
This is the correct way of writing the previous program.

Q10. Find the o/p of this:
main( ) {

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

for(j=0; j<5; j++){ printf(“%d” ,*; a++; } p = a; for(j=0; j<5; j++){ printf(“%d ” ,*p); p++; } }

Ans:

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.

Q11. Find the o/p:
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);

}

Ans:

Q12. Find the o/p:
main() {
int i,j;

for(i=0;i<=10;i++){ j+=5; assert(i<5); }

Ans:

Runtime error: Abnormal program termination.

Explaination:
assert failed (i<5), ,
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.

Q13.What are the files which are automatically opened when a C file is executed?

Ans:

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

Test your C Programming skills Part 2

2011-11-25T18:19:00.000-08:00

Q1. How would you declare a pointer variable ptr so that it can not be used to modify the string it points to (although ptr itself can be changed to point to some other string)?

Ans:

char * const ptr = "Hello";

Q2. How are the string literals are processed by compiler?

Ans:

String literals are allocated in data area that may be read-only.

Q3.
int a[25];
int * p = a + 3;

Which one of the following is equivalent to the above code snippet?
a) int a[25];
int *p = &a[3];

b) int a[25];
int * p = &(a + 3);

c) int a[25];
int * p = (&a)[3];

d) int a[25];
int * p = &a[2];

e) int a[25];
int * p = a[3];

Ans:

Q4. How would you declare a constant pointer to a constant string?
Ans:

const char * const ptr;

Q5. Find the o/p:
int divider(double m, double n)
{
return m/n;
}

main()
{
printf("%d", divider(10/3);
}

Ans:

Q6. Find the o/p:
main(){

     struct xx{

          char name[]="hello";

};

     struct xx *s;

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

}
Ans:

     Compiler Error

     Explanation:
     You should not initialize variables in structure declaration.

Q7. Find the o/p:
void main()
{
int const *p=5;
printf("%d",++(*p));

}
Ans:

Compiler error: Cannot modify a constant value.

Q8. Find the o/p:
# include < stdio.h >
main()
{
     float me = 1.1;
     double you = 1.1;

     if(me==you)
          printf("I love U");

     else
          printf("I hate U");

}

Ans:

I hate U

Q9. Find the o/p:
#include <stdio.h>
main() {
     static int var = 5;
     printf("%d ",var--);
     if(var)
          main();
}

Ans:

5 4 3 2 1

Q10. Find the o/p of this:
main(){

     extern int i;

     i=20;

     printf("%d",i);

}
Ans:

Linker Error : Undefined symbol '_i'

Q11. Find the o/p:
main(){

char *p;

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

}
Ans:

1 4

Q12. Find the o/p:
main(){

     char string[]="Hello World";

     display(string);

}

void display(char *string){

     printf("%s",string);

}
Ans:

Compiler Error : Type mismatch in redeclaration of function display

Q13. What is output?
main(){

     int i=10;

     i=!i>14;

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

}

Ans:

i=0

Test your C Programming skills Part 1

2011-05-26T01:48:00.000-07:00

Q1. Find the o/p:
main()
{
char *ptr="hello";
char ary[]="world";

printf("PTR: &ptr=%x, ptr=%x, *ptr=%x, ptr[0]=%x\n", &ptr,ptr, *ptr, ptr[0]);
printf("ARY: &ary=%x,ary=%x, *ary=%x, ary[0]=%x\n", &ary, ary, *ary, ary[0]);
}

Ans:

PTR: &ptr=bfb86990, ptr=8048524, *ptr=68, ptr[0]=68
ARY: &ary=bfb8698a,ary=bfb8698a, *ary=77, ary[0]=77

Q2. Find the o/p:
main()
{
    struct node {
        char a;
        int b;
        short c;
        struct node next;
    }n1;

printf("Ans1:%d\n", sizeof(n1));

return 0;
}

Ans:

This program will not compiler.
Reason: The compiler can't figure out the space needed for the structure n1.
[kongkond@jorhat ~]$ gcc Q2.c
Q2.c: In function ‘main’:
Q2.c:7: error: field ‘next’ has incomplete type

Q3. Find the o/p:
main()
{
    struct node {
        char a;
        int b;
        short c;
        struct node *next;
    }n1;

printf("Ans1:\n\t node=%x\n \t node+2=%x\n", (char *)&n1, (char *)&n1+2);
printf("Ans2:%d\n", sizeof(n1));

return 0;
}
Ans:

Ans1:
node=bfcc5c94
node+2=bfcc5c96
Ans2:16

Q4. What is the difference between fopen() and open()? When do you choose to use one over the other?

Ans:

FILE * fopen(const char *restrict filename, const char *restrict mode);
--Upon successful completion fopen() returns a FILE pointer. Otherwise, NULL is returned and the global variable errno is set to indicate the error.

int open(const char *pathname, int flags);
int open(const char *pathname, int flags, mode_t mode);
--Given a pathname for a file, open() returns a file descriptor, a small, non-negative integer for use in subsequent system calls (read(2), write(2), lseek(2), fcntl(2), etc.). The file descriptor returned by a successful call will be the lowest-numbered file descriptor not currently open for the process.

Q5. Find the o/p:
main()
{
if(!printf("Hello ")){}
else
printf("World!");
}
Ans:

Hello World!

Q6. Find the o/p:
main()
{
struct i{
int a; //4
double b; //8
char * c; //4
char d[7]; //8
short e; //4
int f; //4
};
printf("sizeof(i)= %d", sizeof(struct i));
}
Ans:

4+8+4+8+4+4=32

Q6.1. Find the o/p:
main()
{
struct i{
int a; //4
double b; //8
char * c; //4
char d[3]; //4
short e; //4
int f; //4
};
printf("sizeof(i)= %d", sizeof(struct i));
}
Ans:

4+8+4+4+4+4=28

Q7. Find the o/p:
# include < stdio.h >
void main()
{
printf(NULL) ;
return ;
}
Ans:

> ./a.out
>

Q8. Find the o/p:
# include < stdio.h >
void main()
{
int i = 0;
printf(i) ;
return ;
}
Ans:

In newer gcc compiler this will not compile. In older ones:
> ./a.out
>

But,
# include < stdio.h >
void main()
{
int i = 0;
printf("%f\n", i) ;
return ;
}

Ans: random number like:
-0.008430

And the result of the following will be like this:
# include < stdio.h >
void main()
{
int i = 0;
printf("%f %d\n", i) ;
return ;
}
Ans:random number like:
-0.008430 134513785

Q9. Find the o/p:
#include <stdio.h>
void main()
{
printf("%s",NULL) ;
return ;
}
Ans:

> ./a.out
(null)

But,
#include <stdio.h>
void main()
{
int i = -10;
printf("%s\n",i) ;
return ;
}
will result in "Segmentation fault" if the value of i is other than zero.

Q10. Find the o/p of this last one:
main()
{
printf("%%",7);
}
Ans:

Try yourself

Q11. How do I declare an array of N pointers to functions returning pointers to functions returning pointers to characters?
Ans:

Well, first you need to know how to declare an array of N items of type T - that's

T foo[N];

Now you need to look at how to declare a pointer to function returning something, say, an object of type S. That's like this:

S (*bar)();

Now assume that S is ``pointer to function returning pointer to char''. We get

(char *) (*)() (*bar)().

So, the whole thing turns out to be (with appropriate parentheses)

(((char)(*))((*)())(((*)((foo)))())([(N)]));

If your compiler complains, break this down into subexpressions.

To call it, just use

foo[i]();

This works because, in C, declaration reflects use, but it's one of those weird distorted mirrors.

Q12. What is asmlinkage and what is it's use?
Ans:

"asmlinkage" is defined, for example, in the header file
"include/asm-i386/linkage.h" as
__attribute__((regparm(0)))
that tells compiler put all function parameters in the stack (i.e.
disables call optimization).
In simple terms, a) you are using assembly code using asmlinkage, and b) all the function parameters are pushed into the stack and optimization is disabled for this purpose.
For more details check this link.

Q13. What is output?
main()
{
printf("%x",-1<<4); }

Ans:

[root@monday kongkon]# gcc 2.c
2.c: In function âmainâ:
2.c:3: warning: incompatible implicit declaration of built-in function âprintfâ
[root@monday kongkon]# ./a.out
fffffff0[root@monday kongkon]#

(1)10=(0000 0000 0000 0001)2
(-1)10=(2's complement of binary 1)=(1's complement of 1 added with 1) =(1111 1111 1111 1110 + 1 )2=(1111 1111 1111 1111)2
Now,
(-1)10 << 4 = (1111 1111 1111 0000)2=(F F F 0)16 Note: In Computer, Negative numbers are represented in 2's complement form, which is 1's complement +1.