如何获取非静态std :: array成员的constexpr`.size()`

问题描述

限时删除！！

鉴于 std :: array< T，N> :: size 是constexpr，在下面的代码段中

Given that std::array<T,N>::size is constexpr, in the snippet below

• Foo1 :: u 不是 static 成员为什么重要?该类型在编译时是已知的，因此它的 size()也是如此.
• Foo2 :: bigger()有什么问题?

• Why does it matter that Foo1::u is not a static member? The type is known at compile time and so is its size().
• What's wrong with Foo2::bigger()?

列表:

// x86-64 gcc 10.1
// -O3 --std=c++20 -pedantic -Wall -Werror

#include <array>
#include <cstdint>

union MyUnion {
    std::array<uint8_t,32> bytes;
    std::array<uint32_t,8> words;
};

struct Foo1 {
    MyUnion u;
    static constexpr size_t length {u.bytes.size()};
        //invalid use of non-static data member 'Foo1::u'
};

struct Foo2 {
    MyUnion u;
    size_t x;
    consteval int8_t length() const { return u.bytes.size(); };
    bool bigger() const { return x > length(); }
        //'this' is not a constant expression
};

我想在 MyUnion 声明中保留std :: array的长度，而不要求助于

I would like to keep std::array length in MyUnion declaration and not resort to

constexpr size_t LEN {32};
union MyUnion {
    std::array<uint8_t,LEN> bytes;
    std::array<uint32_t,LEN/4> words;
};

推荐答案

这些情况有些不同.

在第一个中:

    static constexpr size_t length {u.bytes.size()};
        //invalid use of non-static data member 'Foo1::u'

您正在尝试访问没有对象的非静态数据成员.那不是你能做的.该表达式需要 some Foo1 .在非静态成员函数的上下文中，它隐式为 this-> u ，但是那里仍然需要一些对象.仅有一个例外:您可以编写 sizeof(Foo1 :: u).

You're trying to access a non-static data member without an object. That's just not a thing you can do. There needs to be some Foo1 associated with this expression. In the context of a non-static member function, it'd implicitly be this->u, but there still needs to be some object there. There's only one exception: you're allowed to write sizeof(Foo1::u).

在第二个中:

    consteval int8_t length() const { return u.bytes.size(); };

在这里， this 指针本身不是常量表达式(我们不知道它指向的是什么)，因此通过它访问任何内容都会失败.这是使用未知引用的更广泛的常量表达式的一部分，其使用方式不会真正影响表达式的常量性(请参阅我的博客文章，网址为 constexpr数组大小问题).我最近写了一篇关于该主题的论文 ，该论文主要侧重于参考文献，但 this 是最狭窄的扩展.

Here, the this pointer is not itself a constant expression (we don't know what it points to), so accessing anything through it fails. This is part of a broader case of constant expressions using unknown references in a way that doesn't really affect the constant-ness of the expression (see my blog post on the constexpr array size problem). I recently wrote a paper on this topic which was focused primarily on references, but this is a narrow extension on top of that.

无论哪种方式，目前都无法使用，因为一切都必须是常数.因此，您必须根据您的建议采取一些措施:分别公开您想要的常量.

Either way, for now this can't work either, because everything has to be a constant. So you'll have to resort to something along the lines of what you suggest: expose the constant you want separately.

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