#ifdef with multiple tokens, is this legal?

Question

Today I came across some C++ code that contains an #ifdef clause like this:

#ifdef DISABLE_UNTIL OTHER_CODE_IS_READY
   foo();
#endif

Note the space between "DISABLE_UNTIL" and "OTHER_CODE_IS_READY". Essentially there are two tokens specified in the #ifdef line.

My question is, is this legal C++ code? (g++ compiles it without any errors, and it apparently just ignores the second token). And if it is legal, should the second token have any effect?

Answer 1

[C++11 16.1], [C++11 16.5] and, incidentally, [C99 6.10.1/4] all say that this is invalid.

if-group:
# if constant-expression new-line group_opt
# ifdef identifier new-line group_opt
# ifndef identifier new-line group_opt

Only one identifier is legal.

GCC's own documentation agrees.

My own tests suggest that only the first identifer is accepted, and the second is simply discarded; this may be to ease the implementation, but the standard does require a diagnostic here, so you should see this when you use the -pedantic flag at least^†.

#include <iostream>
using namespace std;

#define A
#define B

int main() {
    #ifdef A
    std::cout << "a ";
    #endif

    #ifdef B
    std::cout << "b ";
    #endif

    #ifdef C
    std::cout << "c ";
    #endif

    #ifdef B C
    std::cout << "bc ";
    #endif

    #ifdef C B
    std::cout << "cb ";
    #endif

    return 0;
}

// Output: "a b bc"
// Note: "cb" *not* output

---

^{† Coliru's installation of GCC emits it with or without -pedantic.}

Answer 2

The syntax you posted is not legal, and the intended meaning is unclear.

Depending on what you hope to accomplish, can you use || or && to combine them?
(of course if this is someone else's code, I'd just reject it as inappropriate / unusable)

#if defined(DISABLE_UNTIL) || defined(OTHER_CODE_IS_READY)
    foo();
#endif