constexpr vs const vs constexpr const [英] constexpr vs const vs constexpr const

问题描述

const-vs-constexpr-on-variables

如果使用double ，那家伙对constexpr的说法是正确的（$code>（或者当然是 float ）。但是，如果将var类型从 double 更改为int，char等整数类型，则一切正常。为什么会发生这种情况？

What the guy says about constexpr is right if double is used (or float of course). However, if you change the var type from double to an integer type like int, char, etc, everything works. Why does that happen?

http://ideone.com/ DAWABE

int main() 
{
    const int PI1 = 3;
    constexpr int PI2 = 3;
    constexpr int PI3 = PI1;  // works
    static_assert(PI1 == 3, "");  // works

    const double PI1__ = 3.0;
    constexpr double PI2__ = 3.0;
    constexpr double PI3__ = PI1__;  // error
    static_assert(PI1__ == 3.0, "");  // error
    return 0;
}

更新：下一行是一个错误，我的意思是 PI3__ = PI1 __

Update: the following line was a mistake, I meant PI3__ = PI1__

constexpr double PI3__ = PI1;  // I meant PI1__

问题：

1. 为什么 const int = 3 是编译时间常数，但 const double = 3.0 不是吗？

1. Why const int = 3 is compile time constant but const double = 3.0 is not?

是否有任何原因我应该使用 constexpr const int val; 超过 constexpr int val ？两者似乎完全一样。

Is there any reason why I should use constexpr const int val; over constexpr int val? They both seem to do exactly the same.

推荐答案

从评论中似乎OP要求使用标准报价，该报价将 const int 定义为编译时常量，但将 const double 定义为编译时常量

From the comments it seems like OP is asking for Standard quote which defines const int as a compile-time constant, but const double as not.

相应的详细信息在5.19 常量表达式中找到。特别是：

The corresponding details are found in 5.19, Constant Expressions. In particular:

...左值到右值转换（4.1），除非应用于非整型或枚举类型的volatile glvalue，它引用具有初始化的
非易失性const对象，并使用常量表达式初始化
...

...an lvalue-to-rvalue conversion (4.1) unless it is applied to a non-volatile glvalue of integral or enumeration type that refers to a non-volatile const object with a preceding initialization, initialized with a constant expression...

int 是整数类型，而 double 不是。

int is an integral type, while double is not.

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