#ifdef具有多个令牌，这合法吗? [英] #ifdef with multiple tokens, is this legal?

问题描述

今天，我遇到了一些C#代码，其中包含这样的#ifdef子句:

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

#ifdef DISABLE_UNTIL OTHER_CODE_IS_READY
   foo();
#endif

请注意"DISABLE_UNTIL"和"OTHER_CODE_IS_READY"之间的空格.基本上，在#ifdef行中指定了两个令牌.

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

我的问题是，这是合法的C ++代码吗? (g ++编译时没有任何错误，并且显然只是忽略了第二个标记).而且如果合法，第二个令牌应该有效吗?

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?

推荐答案

[C++11 16.1]，[C++11 16.5]和[C99 6.10.1/4]都说这是无效的.

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

if-group :
# if 常量表达式 换行符 group _opt
# ifdef 标识符 换行符 group _opt
# ifndef 标识符 换行符 group _opt

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

只有一个标识符是合法的.

Only one identifier is legal.

GCC自己的文档同意.

我自己的测试建议仅接受第一个标识符，而第二个则被丢弃；这可能是为了简化实施，但是该标准确实需要在此处进行诊断，因此，当至少使用-pedantic标志^† 时，应该看到此信息.

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安装的GCC发出它使用或或没有-pedantic .}

---

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

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