在同一个类中使用 constexpr 作为模板参数时出错

2021-12-31 00:00:00 compiler-errors c++ c++11 constexpr

如果我尝试编译以下 C++0x 代码，则会出现错误:

If I try to compile the following C++0x code, I get an error:

template<int n> struct foo { };

struct bar {
    static constexpr int number() { return 256; }

    void function(foo<number()> &);
};

使用 gcc 4.6.1，错误信息是:

With gcc 4.6.1, the error message is:

test.cc:6:27: error: ‘static constexpr int bar::number()’ used before its definition
test.cc:6:28: note: in template argument for type ‘int’

使用 clang 2.8，错误信息是:

With clang 2.8, the error message is:

test.cc:6:20: error: non-type template argument of type 'int' is not an integral
      constant expression
        void function(foo<number()> &);
                          ^~~~~~~~
1 error generated.

如果我将 constexpr 函数移动到基类，它在 gcc 上工作，并在 clang 上给出相同的错误消息:

If I move the constexpr function to a base class, it works on gcc, and gives the same error message on clang:

template<int n> struct foo { };

struct base {
    static constexpr int number() { return 256; }
};

struct bar : base {
    void function(foo<number()> &);
};

是代码错误，还是 gcc 4.6 对 C++0x 实现的限制或错误?如果代码错了，为什么错了，C++11标准的哪些条款说错了?

Is the code wrong, or is it a limitation or bug on gcc 4.6's implementation of C++0x? If the code is wrong, why is it wrong, and which clauses of the C++11 standard say it is incorrect?

推荐答案

在 C++ 中，类的成员函数的内联定义仅在类中的每个声明都被解析后才被解析.因此，在您的第一个示例中，编译器无法在声明 function() 的位置看到 number() 的定义.

In C++, inline definitions of member functions for a class are only parsed after every declaration in the class is parsed. Therefore, in your first example, the compiler can't see the definition of number() at the point where function() is declared.

(clang 的发布版本不支持计算 constexpr 函数，所以你的测试用例都不会在那里工作.)

(No released version of clang has support for evaluating constexpr functions, so none of your testcases will work there.)

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