C Aptitude Questions with Answers Part 8

36)

  #include<stdio.h> 
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).

  Enumerated types can be used in case statements.

37)

  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.

38)

 #define f(g,g2) g##g2 
main() {
   int var12=100; 
   printf("%d",f(var,12)); 
 } 

Answer:
100

39)

 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).

40)

 #include<stdio.h> 
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");