Is it legal to use #elif with #ifdef?

Question

A simple question that Google doesn't help me with. Is it legal in C++ to use #elif clause in the context of #ifdef? It seems to compile and work as expected with all the major compilers in the c++11 mode (MSVC 2015/2017, clang, GCC), but I'm not certain whether it is standard-compliant.

Answer 1

Yes, the grammar allows an #elif after preceding, matching #if, #ifdef or #ifndef:

if-section:
    if-group elif-groups_optelse-group_optendif-line

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

Note that #ifdef(X) is just short for #if defined(X), and #ifndef(X) for #if ! defined(X).

Answer 2

For me, the most important point on this question is actually rosshjb's comment under the question:

@RemyLebeau Yes, we can use #ifdef with #elif. But, If we #define macro with value 0 for #ifdef case, the #ifdef case tests it to true. Otherwise, if we #define macro with value 0 for #elif case, the #elif case test it to false. – rosshjb Jan 19 '20 at 19:40

So if you have a block like:

#ifdef __linux__
  <some Linux code here>
#elif _WIN32
  <some Windows code here>
#endif

Then the second test is significantly different from the first - the first is checking whether __linux__ is defined at all, where the second is checking that the symbol _WIN32 evaluates to true. For many cases it will behave the same, but it is not guaranteed to do so.

The full equivalent is actually:

#ifdef __linux__
  <some Linux code here>
#elif defined(_WIN32)
  <some Windows code here>
#endif

Which is probably not obvious to everyone.

Using Kerrick SB's answer, you can also write the same #if statement as:

#if defined(__linux__)
  <some Linux code here>
#elif defined(_WIN32)
  <some Windows code here>
#endif

Which makes it more obvious that the defined is the same for both the #if and the #elif