How the output is 10 but not 15?

Question

#include <stdio.h>
#define prod(i, j) i * j
int main()
{
int x = 3, y = 4;
printf("%d", prod(x + 2, y - 1));
return 0;
}

Answer 1

x+2*y-1=3+2*4-1=3+8-1=10

Answer 2

NIKHIL JOSHI (´✪ω✪`) Note: Pre-processor directives just replace the text. Do not compare it with functions.

Answer 3

Martin Taylor what is the benefit of this method(replacing oprations by preprocessor) comparing to functions?

Answer 4

Mehran I use macros when trying to make small functions, the job of whom can be done by simple text substitution, which is better in my opinion.

Answer 5

If the preprocessor directive,

#define prod(i, j) i * j

were written the safe way, with parentheses, as,

#define prod(i, j) ((i)*(j))

the preprocessor would replace

printf("%d", prod(x + 2, y - 1));

with

printf("%d", ((x + 2) * (y - 1)));

and the result would be 15.

Answer 6

Bcuz of precedence of operator