I do not understand the behaviour of #define macro in C++

Question

I need to understand how this code works:

#define foo1( a ) (a * a)              // How does this work?
inline int foo2(         


        

Answer 1

Well, it's really simple.

foo1( 1 + 2 )

will turn into:

( 1 + 2 * 1 + 2 )

which is actually:

1 + 2 + 2 = 5

This is how macros work.