constexpr对非const对象的引用 [英] constexpr reference to non-const object

问题描述

是否可以将非常量引用声明为constexpr?示例代码:

Is it permitted to declare a non-const reference as constexpr? Example code:

int x = 1;
constexpr int& r = x;

这被gcc和clang接受(我尝试了两个当前和过去的版本，回到C ++ 11，都接受了).但是我认为它不应该被接受，因为C ++ 14 [dcl.constexpr/9]说:

This is accepted by gcc and clang (I tried several current and past versions of both, back to C++11, and all accepted it). However I think it should not be accepted because C++14 [dcl.constexpr/9] says:

如果在引用声明中使用constexpr说明符，则每个完整的- 初始化程序中出现的表达式应为常量表达式

if a constexpr specifier is used in a reference declaration, every full- expression that appears in its initializer shall be a constant expression

和x不是常数表达式.

[dcl.constexpr]的最新C ++ 17草案中的语言已更改，甚至不再明确提及constexpr引用，我无法说出它试图说的话.他们.

The language in the latest C++17 draft of [dcl.constexpr] changed and doesn't even mention constexpr references explicitly any more, I can't make head nor tail of what it is trying to say about them.

推荐答案

假定x具有静态存储期限，则左值表达式x是一个完全有效的常数表达式.

Assuming that x has static storage duration, the lvalue expression x is a perfectly valid constant expression.

如果在需要prvalue的上下文中使用x，这会导致从左值到右值的转换被应用，那么生成的prvalue表达式-称为TO_RVALUE(x)-将不是常量表达式，原因很明显.但是在引用绑定的情况下，没有这种转换.

If you use x in a context that requires a prvalue, which causes the lvalue-to-rvalue conversion to be applied to it, then the resulting prvalue expression - call it TO_RVALUE(x) - would not be a constant expression, for obvious reasons. But in the case of reference binding, there is no such conversion.

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