C programming Language quiz with solution

1)

       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.

2) 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.

3) 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.

4) #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.

5) 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".

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