#define的奇怪行为

Question

I have this code in C++:

#include <string>
#include <iostream>

int const foo = 1;
int const bar = 0;

#define foo bar
#define bar foo

int main()
{
  std::cout << foo << std::endl;
  std::cout << bar << std::endl;
}

It produces this output:

bash-3.2$ ./a.out
1
0

I do not understand why this is the output.

Answer 1

Macros will never expand recursively.

When you write foo , it first expands to bar , and then since bar is a macro it expands back to foo . While foo is a macro, because macros can't be recursive it will not be expanded. And then evaluating foo yields its value: 1.

The same goes for bar .

See this: http://gcc.gnu.org/onlinedocs/cpp/Self_002dReferential-Macros.html#Self_002dReferential-Macros

And ISO/IEC 14882:2003(E) 16.3.4 Rescanning and further replacement section of the standard. (see comments for more details)