What’s going to happen when we compile and run the following C program snippet?
#include "stdio.h" int main()
{ int a = 10;
int b = 15;
printf("=%d",(a+1),(b=a+2));
printf(" %d=",b);
return 0;
} |
(A) =11 15=
(B) =11 12=
(C) Compiler Error due to (b=a+2) in the first printf().
(D) No compile error but output would be =11 X= where X would depend on compiler implementation.
Answer: (B)
Explanation: As per C standard C11, all the arguments of printf() are evaluated irrespective of whether they get printed or not. That’s why (b=a+2) would also be evaluated and value of b would be 12 after first printf(). That’s why correct answer is B.
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