C++ define expression evaluation

Question

Suppose we have this expression:

#define cube(x) x * x * x

And then we call it:

int n = 3, v;
v = cube(n + 1);   // v = 10
v = cube((n + 1)); // v = 64
v = cube(n);       // v = 27

So the question is: why first operation do not make v = 64?

Answer 1

Macros are not evaluated (in the sense of the common interpretation of evaluation), they are expanded at compile time.

Before the file is compiled, there is another program called the C Preprocessor that replaces the macro invocation literally/textually and prepares the file for actual compilation, so for your macro

#define cube(x) x * x * x when you do this

This

v = cube(n + 1);

is replaced with this (expaned is the correct term)

v = n + 1 * n + 1 * n + 1;
// Simplifies to
v = n + n + n + 1;
// and again
v = 3 * n + 1;

which for n = 3 gives you 10 exactly the observed result.

Note, that when you add parentheses

v = cube((n + 1));

then, the expansion is

v = (n + 1) * (n + 1) * (n + 1);

which is what you would expect cube() to do, so prevent this you should redefine your macro like this

#define cube(x) ((x) * (x) * (x))

If you are using gcc try

gcc -E source.c

and check the result to verify how the macro was expanded.