c++ - #define 可以被赋予定义的结果吗？

Question

以下代码的结果是用 C 或 C++ 定义的吗？

#define FOO
#define BAR defined(FOO)
#if BAR
    int x = 1;
#else
    int x = 2;
#endif

Answer 1

使用已定义作为条件指令中的扩展被视为未定义行为。

• 来自 C99 标准:6.10.1/4 | C++11 Standard Last Working Draft(n4296) 16.1/4

If the token defined is generated as a result of this replacement process or use of the defined unary operator does not match one of the two specified forms prior to macro replacement, the behavior is undefined.

• 来自 C99 标准:6.10.8/4 | C++11 Standard Last Working Draft(n4296) 16.8/4

None of these macro names, nor the identifier defined, shall be the subject of a #define or a #undef preprocessing directive.

• GNU - CPP 4.2.3

If the defined operator appears as a result of a macro expansion, the C standard says the behavior is undefined. GNU cpp treats it as a genuine defined operator and evaluates it normally. It will warn wherever your code uses this feature if you use the command-line option ‘-pedantic’, since other compilers may handle it differently.