How this square function works?

Question

Can someone explain why the result of this function is 9.

#define square(x) x*x
int main() {
printf("%d", square(4+1));
return 0;
}

Answer 1

In this case square(x) means that the x will be replaced by whatever you put it. It will not be processed.

Hence, what happens is the compiler treats it like 4+1*4+1, because x is treated as 4+1.

And if you know your order of operations you know the answer is 4+4+1=9.

This is a trouble with macros. To fix this, do:

#define square(x) (x)*(x)

And square(4+1) will output 5² correctly, which is 25.

Answer 2

It's not doing (4+1)*(4+1) but rather 4+1*4+1 == 4 + 4 + 1 == 9.

Answer 3

Thanks for the answers and very good explanation from Prometheus.