结构化绑定:当某物看起来像参考并且行为类似于参考，但不是参考时

问题描述

昨天我见过.
我们可以总结如下.考虑下面的示例代码:

Yesterday I've seen an interesting question here on SO about structured binding.
We can sum up it as it follows. Consider the example code below:

#include <tuple>
#include <type_traits>

int main() {
    auto tup = std::make_tuple(1, 2);
    auto & [ a, b ] = tup;
    // the following line won't compile for a isn't a reference
    // static_assert(std::is_reference_v<decltype(a)>);
}

在这种情况下，decltype(a)是int(可能是由于此项目符号(工作草案):

In this case decltype(a) is int (probably) because of this bullet (working draft):

此处是@Curious在感兴趣的评论中提供的wandbox的摘要.它表明实际上a不是参考，仅此而已.
到目前为止，对于最初的问题如此好，OP询问为什么它是int而不是int &，并且标准说看起来是可以接受的答案.

Here is a snippet on wandbox provided by @Curious in the comments for those that are interested. It shows that actually a isn't a reference, nothing more.
So far so good for the original question, OP asked why it was int instead of int & and the standard says that looked like an acceptable answer.

无论如何，我想知道委员会为何如此决定.归根结底，a是指元组中的元素，我可以通过a修改该元素.换句话说，a的声明看起来像是引用中的一个，它的行为类似于引用，但不是引用.

Anyway, I'd like to know why the committee decided so. At the end of the day, a refers to an element in the tuple and I can modify that element through a. In other terms, the declaration of a looks like the one of a reference, it behaves similarly to a reference but it's not a reference.

我可以接受这一点，但是我想知道背后的原因是什么.为什么decltype(a)不能简单地是int &?亵渎者可以理解的有意义的原因吗?

I can live with this, but I'd like to know what are the reasons behind that. Why decltype(a) cannot be simply int &? Is there a meaningful reason that a profane can understand?

推荐答案

我昨天写了这个:

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