C Technical Questions Part 5

C Technical Questions

1.) 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]);

}

Ans:

mmmm

aaaa

nnnn

Explanation:

Here the for loop will execute until a null character is reached. s[i], i[s], *(i+s), *(s+i) all indicates i th element of array s.

2. main()

{

            char *p;

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

}

Ans:

1 2

Explanation:

sizeof(*p) returns the size of the character pointed by p. sizeof(p) returns the size of pointer p. Note:// The size of normal pointers are always 2.

3.) main()

{

              printf("%x",-1<<4);

}

Ans:

0xfff0

Explanation:

The binary representation of -1  and corresponding hexa representation is.

11111111 11111111  

F       F       F      F

11111111 11111111   << 4 gives

1111 1111 1111 0000

F        F       F       0

Hence the result

4.)  main()

{

            int c=- -2;

            printf("c=%d",c);

}

Ans:

 2

Explanation:

Use of two unary operators minus operator cancel the effect each other.

Note:// the space between - - is necessary otherwise it will be treated as decrement operator.

5.) main()

{

 char *p;

 p="Hello";

 printf("%c\n",*&*p);

}

//here*&p==&*p

Ans:

H

Explanation :

*p specifies value at address p.

&*p  equal to &(*p) specifies address of value pointed by p, ie nothing other than address stored in p.

*&*p equal to *(&(*p)) specifies value at address of value pointed by p, ie nothing other than *p which is value ‘H’

6.) main()

            {

            char *str1="abcd";

            char str2[]="abcd";

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

            }

Ans:

2 5 5

Explanation:

sizeof(str1),sizeof(str2),sizeof("abcd")

sizeof(str1) returns the size of pointer variable str1 which is 2 (size of a normal pointer is always 2). sizeof(str2) returns the size of array str2 which is 5*sizeof(char) hence 5 (the fifth element of array is the null character. sizeof("abcd") returns size of string constant “abcd” which is again 5*sizeof(char).(there will always a null character appended at the end of the string constant).

7. #include<stdio.h>

main()

  {

    register i=5;

    char j[]= "hello";                    

    printf("%s  %d",j,i);

}

Ans:

hello 5

Explanation:

The keyword register specifies the value at i will be stored in memory register. register is a storage class, if data type is not mentioned after a storage class or qualifier by default it will be treated as int.

8.) void main()

{

            int i;

            char a[]="\0";

            if(printf("%s\n",a))

            printf("Ok here \n");

            else

            printf("Forget it\n");

}

Ans:

Ok here

Explanation:

  Printf returns the number of characters it successfully printed. Here printf prints one newline character hence the result.

9. What is the output of the program given below

main()

    {

       signed char i=0;

       for(;i>=0;i++) ;

       printf("%d\n",i);

    }

Ans :

-128

Explanation:

Signed characters will take values from -128 to +127 in cyclic manner. Here first the value 0 is assigned to i then the i is continuously incremented in the increment section of the loop. When the value of i reaches 127, when it again incremented i will take value -128 and the test condition fails and loop terminates. Hence the result

10.) main()

      {

       char i=0;

       for(;i>=0;i++) ;

       printf("%d\n",i);

}

Ans :

 -128

Explanation:

Works in the same manner as above. Normally when no qualifiers are used by default data types are taken as signed.

11.) void main()

      {

      if(~0 == (unsigned int)-1)

      printf(“You can answer this if you know how values are represented in memory”);

      }

      Ans :

      you can answer this if you know how values are represented in memory.

      Explanation:

      As I mentioned in the previous questions variables will take values within its range in a cyclic manner. Unsigned int will take values between 0 to65535 and signed between -32768 to +32767. The unsigned equivalent of -1 is 1. Hence the result.

         

12. main()

{

int i = 3;

for (;i++=0;) printf(“%d”,i);

}

Ans :

L value required error.

Explanation:

 i++ will return athe value 3 and is impossible to assign a value to a constant.

13. main()

{

 int i=5,j=10;

i=i&=j&&10;

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

}

Ans :

 1 10

Explanation:

i=i&=j&&10; statement will be expanded as i=i=i&j&&10; hence the result.

14.)

#define DIM( array, type) sizeof(array)/sizeof(type)

main()

{

int arr[10];

printf(“The dimension of the array is %d”, DIM(arr, int));   

}

Ans:

10

Explanation:

Preprocessor will expand the macro as sizeof(arr)/sizeof(int) which is 20/2, hence the result.

15. int DIM(int array[])

{

return sizeof(array)/sizeof(int );

}

main()

{

int arr[10];

printf(“The dimension of the array is %d”, DIM(arr));   

}

Ans:

The  dimension of the array is 1

Explanation:

An array can be passed to a function only as a pointer and size of pointer is always 2 hence the result.

16. #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:

Error

Explanation:

Arr1 and arr2 is not defined anywhere.