c++ - 为什么右值引用整数合法

Question

关于C++ 11，这可能是一个愚蠢的问题，但这确实让我感到困扰。
据我了解，右值引用也是一个引用，这意味着它将引用指向到某个变量，就像引用一样。
例如，

const int &ref = 1;

引用ref指向无法修改的纯右值1，这就是为什么编译器迫使我们使用const的原因。
另一个例子，

Bar&& GetBar()
{
    Bar b;
    return std::move(b);
}

此函数将返回一个悬空的引用，因为b写入后return被破坏了。
一言以蔽之，右值引用就是一个引用。
现在我很困惑。请检查以下代码:

int &&rref = 1;

如果右值引用也是一个引用，那么现在rref 将指向纯右值1，根据我的理解，该值不应该编译，因为如果它是可编译的，那么我执行rref = 2怎么办？这是否意味着纯右值已更改:1变为2？
但是gcc告诉我它是可编译的...
为什么？为什么我们不需要const int &&rref = 1？

Answer 1

cppreference的报价

The lifetime of a temporary object may be extended by binding to a const lvalue reference or to an rvalue reference (since C++11), see reference initialization for details.

带有指向其他详细信息的链接here

Whenever a reference is bound to a temporary or to a subobject thereof, the lifetime of the temporary is extended to match the lifetime of the reference, with the following exceptions:

然后继续列出一些异常(exception)，例如

a temporary bound to a return value of a function in a return statement is not extended: it is destroyed immediately at the end of the return expression. Such function always returns a dangling reference.

a temporary bound to a reference member in a constructor initializer list persists only until the constructor exits, not as long as the object exists. (note: such initialization is ill-formed as of DR 1696). (until C++14)

a temporary bound to a reference parameter in a function call exists until the end of the full expression containing that function call: if the function returns a reference, which outlives the full expression, it becomes a dangling reference.

a temporary bound to a reference in the initializer used in a new-expression exists until the end of the full expression containing that new-expression, not as long as the initialized object. If the initialized object outlives the full expression, its reference member becomes a dangling reference. (since C++11)

a temporary bound to a reference in a reference element of an aggregate initialized using direct-initialization syntax (parentheses) as opposed to list-initialization syntax (braces) exists until the end of the full expression containing the initializer. (since C++20)

所以推理

const int &ref = 1;

有效，因为我们使用的const不正确。编译器实际上是在扩展临时对象的生存期，以匹配引用的生存期。因此，使用右值引用执行相同的操作同样有效。
另一方面

int &ref = 1;

因为1不是左值，所以没有任何意义。这就是为什么我们需要const或右值引用的原因。