使用引用的constexpr静态成员作为模板参数 [英] Using a constexpr static member of a reference as template argument

问题描述

我试图弄清楚是GCC还是Clang对C ++ 17标准的解释不同/错误.

I'm trying to figure out whether GCC or Clang interpret the C++17 standard differently / wrong here.

这是我的代码，使用GCC 8进行编译，但不使用Clang 6:

This is my code, which does compile using GCC 8, but not using Clang 6:

struct BoolHolder {
    constexpr static bool b = true;
};

template<bool b>
class Foo {};

int main() {
    BoolHolder b;
    Foo<b.b> f; // Works

    BoolHolder & br = b;
    Foo<br.b> f2; // Doesn't work
}

我想知道为什么会这样.显然，b.b是有效的constexpr(或第一个Foo<b.b>无效). br.b不是有效的constexpr吗?为什么?对象或引用本身应该与它无关，因为我们在这里访问静态constexpr成员，对吧?

I wonder why that is. Obviously, b.b is a valid constexpr (or the first Foo<b.b> wouldn't be valid). Is br.b not a valid constexpr? Why? The object or the reference itself should have nothing to do with it, since we're accessing a static constexpr member here, right?

如果这实际上不是有效的C ++ 17，那么即使GCC都没有警告我(即使我启用了-Wall -Wextra -pedantic)这一事实，也应该被视为错误吗?

If this is really not valid C++17, should the fact that GCC doesn't even warn me (even though I enabled -Wall -Wextra -pedantic) be considered a bug?

推荐答案

C语正确.可以这么说，引用以常量表达式急切"地求值. [expr.const]/2.11:

Clang is correct. References are evaluated "eagerly" in constant expressions, so to speak. [expr.const]/2.11:

表达式e是核心常量表达式，除非求值 的e，按照抽象机的规则，将求出一个 以下表达式之一:

An expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

• [...]
• 一个 id-expression ，它引用引用类型的变量或数据成员，除非引用具有前面的初始化，并且 任何一个
  • 使用常量表达式或
  进行初始化
  • 其寿命始于对e的评估；
  • [...]
  • an id-expression that refers to a variable or data member of reference type unless the reference has a preceding initialization and either
    • it is initialized with a constant expression or
    • its lifetime began within the evaluation of e;

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