C preprocessors and order of operations

Question

I'm learning C, but I do not understand this:

#define square(x) x*x
a = square(2+3) //a = 11

When this is run, why does a end up being 11?

Answer 1

It expands to 2+3*2+3, which is equivalent to 2+(3*2)+3. Use parentheses to fix it:

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

Now try it with square(x++) and you'll run into more problems (undefined behavior). Avoid doing this as a macro if you can.

Answer 2

square(2+3) expands to 2+3*2+3 which is equivalent to 2+(3*2)+3 [* has higher precedence than +]

On gcc you can use -E option to see what your preprocessor generates

C:\Users\SUPER USER>type a.c
#define square(x) x*x

int main()
{
   a = square(2+3); //a = 11
}

C:\Users\SUPER USER>gcc -E a.c
# 1 "a.c"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "a.c"


int main()
{
   a = 2+3*2+3;
}

Remedy

Try this

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