Output of C Program | Set 17

Predict the output of following C programs.

Question 1

#include<stdio.h>   #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<stdio.h> #define f(g,g2) g##g2 int 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, afterpreprocessing, the main function becomes as follows, and prints 100.

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

Question 3

#include<stdio.h> int 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.

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