为什么constexpr必须是静态的? [英] Why constexpr must be static?

问题描述

试图使用constexpr属性创建结构的成员而不是静态的尝试会导致编译器错误(请参见下文).这是为什么?对于单个常数值，我将在内存中保留该值，直到终止程序，而不仅仅是结构范围?我应该回到使用宏吗?

An attempt to create a member of a struct with constexpr attribute without being static result in a compiler error(see below). Why is that? for a single constant value will I have this value in memory until program is terminatted instead of just scope of struct? should I back to use a macro?

struct foo
{
  constexpr int n = 10;
  // ...
};

error: non-static data member cannot be constexpr; did you intend to make it static?

推荐答案

我不知道官方的合理性.但是可以肯定会导致混乱.我无法理解，非静态数据成员成为 constexpr 意味着什么.您能执行以下操作吗?

I don't know the official rational. But surely it could lead to confusion. I, for one, can't see what it means for a non-static data member to be constexpr. Are you able to do the following?

struct foo {
  constexpr int n = 10;
  constexpr foo() { }
  constexpr foo(int n):n(n) { } // overwrite value of n
};

还是这意味着初始化程序必须始终是常量，即不允许您编写以上内容(因为 n 不是常量/可能是非常量的)，而是可以允许

Or does it mean that the initializer must be constant always, i.e you are not allowed to write the above (because n is not constant/could potentially non-constant) but allowed to say

foo f = { 10 };

constexpr int n 只是格式错误，而不是隐式地为 static 的规则对我来说似乎很好，因为IMO尚不清楚其语义.

The rule that constexpr int n is simply ill-formed rather than being implicitly static seems good to me, as its semantics would not be clear IMO.

这篇关于为什么constexpr必须是静态的?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！