C INTERVIEW QUESTIONS - 5 :   

PREDICT THE OUTPUT OR ERROR(S) FOR THE FOLLOWING C SNIPPETS:

1.

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.

2.

main()

{

char not;

not=!2;

printf("%d",not);

}

Answer:

0

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.

3.

#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.

4.

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".

5.

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.

6.

#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.

7.

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.

8.

main()

{

int *j;

{

int i=10;

j=&i;

}

printf("%d",*j);

}

Answer:

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.

9.

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.

10.

#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.

11.

#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.

12.

#include‹stdio.h›

main()

{

register i=5;

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.

13.

main()

{

int i=5,j=6,z;

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

}

Answer:

11

Explanation:

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

14.

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:

2

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.

15.

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)

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.