Explaination for #define function.

Question

#define sqr(X) X*X
int main()
{
printf("%d",sqr(3+1));
}

Ans = 9
I don't know that how it is executing.
please help me anyone here.

Answer 1

You actually get 7.

Explanation:
3+1 is actually pasted into X, since macros do work like text substitution. So it becomes:
sqr(3+1)=3+1*3+1=7

Solution:
wrap every argument of the macro with parentheses, like:
#define sqr(X) (X)*(X)
In that case it will be:
(3+1)*(3+1)=4*4=16

Answer 2

Sir thanks for replay. I tried this program in challenge.But I got that answer as "9".
I just reported as "wrong answer".

Answer 3

You're very welcome :)