Output of C Program | Set 17

• Difficulty Level : Medium
• Last Updated : 31 Jul, 2018

Predict the output of following C programs.

Question 1

 #include  #define R 10#define C 20  int main(){   int (*p)[R][C];   printf("%d",  sizeof(*p));   getchar();   return 0;}

Output: 10*20*sizeof(int) which is “800” for compilers with integer size as 4 bytes.
The pointer p is de-referenced, hence it yields type of the object. In the present case, it is an array of array of integers. So, it prints R*C*sizeof(int).
Thanks to Venki for suggesting this solution.

Question 2

 #include#define f(g,g2) g##g2int main(){   int var12 = 100;   printf("%d", f(var,12));   getchar();   return 0;}

Output: 100
The operator ## is called “Token-Pasting” or “Merge” Operator. It merges two tokens into one token. So, after preprocessing, the main function becomes as follows, and prints 100.

 int main(){   int var12 = 100;   printf("%d", var12);   getchar();   return 0;}

Question 3

 #includeint main() {   unsigned int x = -1;   int y = ~0;   if(x == y)      printf("same");   else      printf("not same");   printf("\n x is %u, y is %u", x, y);   getchar();   return 0;}

Output: “same x is MAXUINT, y is MAXUINT” Where MAXUINT is the maximum possible value for an unsigned integer.
-1 and ~0 essentially have same bit pattern, hence x and y must be same. In the comparison, y is promoted to unsigned and compared against x. The result is “same”. However, when interpreted as signed and unsigned their numerical values will differ. x is MAXUNIT and y is -1. Since we have %u for y also, the output will be MAXUNIT and MAXUNIT.
Thanks to Venki for explanation.