根据 c++ 标准，当 pushing/popping 宏定义中的宏时，行为是什么
What is the behavior when pushing/popping a macro inside that macro definition according to c++ standard
我刚才回答了一个关于附加到 C 宏的问题
在 Can I append to a preprocessor macro?
正如答案所说，以下内容在 clang 和 g++ 中都有效，但在 msvc 中无效
#define pushfoo _Pragma("push_macro(\"foo\")") //for convenience
#define popfoo _Pragma("pop_macro(\"foo\")") //I tried __pragma and __Pragma for msvc as well
#define foo 1
pushfoo //push the old value
#undef foo //so you don't get a warning on the next line
#define foo popfoo foo , 2 //append to the previous value of foo
pushfoo
#undef foo
#define foo popfoo foo , 3
pushfoo
#undef foo
#define foo popfoo foo , 4
foo //this whole list will expand to something like popfoo foo popfoo foo popfoo foo , 4
//which will in turn expand to 1 , 2 , 3 , 4
foo //the second time this will expand to just 1
虽然这看起来确实有效，但我想知道这是否恰好适用于 gcc
和 clang，但 msvc 也是（或仅）正确的，或者如果它们表现出的行为是强制的行为
由语言规范。如果 C 和 C++ 有所不同，我对它们都感兴趣（主要是对
最新版本，写作时的 C18 和 C++17）
What is the behavior when pushing/popping a macro inside that macro definition according to c++ standard?
在 C 和 C++ 中，#pragma 和 _Pragma 的结果是实现定义的，因此标准不管理它们的行为。参见 C11 6.10.6p1:
A preprocessing directive of the form
# pragma pp-tokensopt new-line
where the preprocessing token STDC does not immediately follow pragma in the directive (prior to any macro replacement)174) causes the implementation to behave in an implementation-defined manner. The behavior might cause translation to fail or cause the translator or the resulting program to behave in a non-conforming manner. Any such pragma that is not recognized by the implementation is ignored.
和6.10.9涵盖_Pragma。
和最新的 C++ 草案 [cpp.pragma]:
A preprocessing directive of the form
# pragma pp-tokensopt new-line
causes the implementation to behave in an implementation-defined manner.*
The behavior might cause translation to fail or cause the translator or the resulting program to behave in a non-conforming manner.
Any pragma that is not recognized by the implementation is ignored.
我刚才回答了一个关于附加到 C 宏的问题 在 Can I append to a preprocessor macro?
正如答案所说，以下内容在 clang 和 g++ 中都有效，但在 msvc 中无效
#define pushfoo _Pragma("push_macro(\"foo\")") //for convenience
#define popfoo _Pragma("pop_macro(\"foo\")") //I tried __pragma and __Pragma for msvc as well
#define foo 1
pushfoo //push the old value
#undef foo //so you don't get a warning on the next line
#define foo popfoo foo , 2 //append to the previous value of foo
pushfoo
#undef foo
#define foo popfoo foo , 3
pushfoo
#undef foo
#define foo popfoo foo , 4
foo //this whole list will expand to something like popfoo foo popfoo foo popfoo foo , 4
//which will in turn expand to 1 , 2 , 3 , 4
foo //the second time this will expand to just 1
虽然这看起来确实有效，但我想知道这是否恰好适用于 gcc 和 clang，但 msvc 也是（或仅）正确的，或者如果它们表现出的行为是强制的行为 由语言规范。如果 C 和 C++ 有所不同，我对它们都感兴趣（主要是对 最新版本，写作时的 C18 和 C++17）
What is the behavior when pushing/popping a macro inside that macro definition according to c++ standard?
在 C 和 C++ 中，#pragma 和 _Pragma 的结果是实现定义的，因此标准不管理它们的行为。参见 C11 6.10.6p1:
A preprocessing directive of the form
# pragma pp-tokensopt new-linewhere the preprocessing token STDC does not immediately follow pragma in the directive (prior to any macro replacement)174) causes the implementation to behave in an implementation-defined manner. The behavior might cause translation to fail or cause the translator or the resulting program to behave in a non-conforming manner. Any such pragma that is not recognized by the implementation is ignored.
和6.10.9涵盖_Pragma。
和最新的 C++ 草案 [cpp.pragma]:
A preprocessing directive of the form
# pragma pp-tokensopt new-linecauses the implementation to behave in an implementation-defined manner.* The behavior might cause translation to fail or cause the translator or the resulting program to behave in a non-conforming manner. Any pragma that is not recognized by the implementation is ignored.