online Placement Question of c programming languge

1) #include

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. 

2)

main()

{

int i;

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

}

Answer:-

1

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. 

3)

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

main()

{

int var12=100;

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

}

Answer:-

100 

4)

main()

{

int i=0;

 

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

printf("%d",i);

}

Answer:-

1

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).

5)

#include

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:-

M

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");

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