constexpr 使用静态函数初始化静态成员

2021-12-18 00:00:00 gcc g++ c++ constexpr static-members

我想要一个从 constexpr 函数计算的 constexpr 值(即编译时常量).而且我希望这两个范围都在类的命名空间内，即静态方法和类的静态成员.

I want a constexpr value (i.e. a compile-time constant) computed from a constexpr function. And I want both of these scoped to the namespace of a class, i.e. a static method and a static member of the class.

我首先用(对我来说)显而易见的方式写了这个:

I first wrote this the (to me) obvious way:

class C1 {
  constexpr static int foo(int x) { return x + 1; }
  constexpr static int bar = foo(sizeof(int));
};

g++-4.5.3 -std=gnu++0x 说:

error: ‘static int C1::foo(int)’ cannot appear in a constant-expression
error: a function call cannot appear in a constant-expression

g++-4.6.3 -std=gnu++0x 抱怨:

error: field initializer is not constant

第二次尝试

好吧，我想，也许我必须把东西移出类体.所以我尝试了以下方法:

Second attempt

OK, I thought, perhaps I have to move things out of the class body. So I tried the following:

class C2 {
  constexpr static int foo(int x) { return x + 1; }
  constexpr static int bar;
};
constexpr int C2::bar = C2::foo(sizeof(int));

g++-4.5.3 将编译它而不会抱怨.不幸的是，我的其他代码使用了一些基于范围的 for 循环，所以我必须至少有 4.6.现在，我仔细查看了 支持列表，看来 constexpr 也需要 4.6.使用 g++-4.6.3 我得到

g++-4.5.3 will compile that without complaints. Unfortunately, my other code uses some range-based for loops, so I have to have at least 4.6. Now that I look closer at the support list, it appears that constexpr would require 4.6 as well. And with g++-4.6.3 I get

3:24: error: constexpr static data member ‘bar’ must have an initializer
5:19: error: redeclaration ‘C2::bar’ differs in ‘constexpr’
3:24: error: from previous declaration ‘C2::bar’
5:19: error: ‘C2::bar’ declared ‘constexpr’ outside its class
5:19: error: declaration of ‘const int C2::bar’ outside of class is not definition [-fpermissive]

这对我来说听起来很奇怪.这里的constexpr"有什么不同?我不想添加 -fpermissive 因为我更喜欢严格检查我的其他代码.将 foo 实现移到类体之外没有明显的效果.

This sounds really strange to me. How do things "differ in constexpr" here? I don't feel like adding -fpermissive as I prefer my other code to be rigurously checked. Moving the foo implementation outside the class body had no visible effect.

有人可以解释一下这里发生了什么吗?我怎样才能实现我正在尝试做的事情?我主要对以下类型的答案感兴趣:

Can someone explain what is going on here? How can I achieve what I'm attempting to do? I'm mainly interested in answers of the following kinds:

  • 一种在 gcc-4.6 中实现此功能的方法
  • 观察到后来的 gcc 版本可以正确处理其中一个版本
  • 一个指向规范的指针，根据该规范，我的至少一个构造应该工作，这样我就可以让 gcc 开发人员在实际让它工作时出错
  • 根据规范，我想要的信息是不可能的，最好能说明此限制背后的基本原理
  • A way to make this work in gcc-4.6
  • An observation that later gcc versions can deal with one of the versions correctly
  • A pointer to the spec according to which at least one of my constructs should work, so that I can bug the gcc developers about actually getting it to work
  • Information that what I want is impossible according to the specs, preferrably with some insigt as to the rationale behind this restriction

也欢迎其他有用的答案，但可能不会那么容易被接受.

Other useful answers are welcome as well, but perhaps won't be accepted as easily.

推荐答案

标准要求(第 9.4.2 节):

The Standard requires (section 9.4.2):

可以在类定义中使用 constexpr 说明符声明文字类型的 static 数据成员；如果是这样，它的声明应指定一个 brace-or-equal-initializer 其中每个作为 assignment-expression 的 initializer-clause 都是一个常量表达式.

A static data member of literal type can be declared in the class definition with the constexpr specifier; if so, its declaration shall specify a brace-or-equal-initializer in which every initializer-clause that is an assignment-expression is a constant expression.

在您的第二次尝试"和 Ilya 的回答中的代码中，声明没有 brace-or-equal-initializer.

In your "second attempt" and the code in Ilya's answer, the declaration doesn't have a brace-or-equal-initializer.

你的第一个代码是正确的.不幸的是 gcc 4.6 不接受它，而且我不知道在哪里可以方便地尝试 4.7.x(例如 ideone.com 仍然停留在 gcc 4.5 上).

这是不可能的，因为不幸的是，标准禁止在类完成的任何上下文中初始化静态 constexpr 数据成员.9.2p2 中brace-or-equal-initializers 的特殊规则仅适用于非静态 数据成员，但这是静态的.

This isn't possible, because unfortunately the Standard precludes initializing a static constexpr data member in any context where the class is complete. The special rule for brace-or-equal-initializers in 9.2p2 only applies to non-static data members, but this one is static.

最可能的原因是 constexpr 变量必须在成员函数体内部作为编译时常量表达式可用，所以变量初始化器完全在函数体之前定义 -- 这意味着函数在初始化器的上下文中仍然是不完整的(未定义的)，然后这个规则开始生效，使得表达式不是一个常量表达式:

The most likely reason for this is that constexpr variables have to be available as compile-time constant expressions from inside the bodies of member functions, so the variable initializers are completely defined before the function bodies -- which means the function is still incomplete (undefined) in the context of the initializer, and then this rule kicks in, making the expression not be a constant expression:

在constexpr 函数或constexpr<的定义之外调用未定义的constexpr 函数或未定义的constexpr 构造函数/code> 构造函数；

an invocation of an undefined constexpr function or an undefined constexpr constructor outside the definition of a constexpr function or a constexpr constructor;

考虑:

class C1
{
  constexpr static int foo(int x) { return x + bar; }
  constexpr static int bar = foo(sizeof(int));
};

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