为什么可以使用类型别名定义对引用的引用？

Question

既然不能定义对引用的引用，为什么这段代码第8行没有报错呢？我的理解是，上述声明等同于 double &&ref2 = value; 这应该会引发错误。

#include <iostream>
using namespace std;
int main()
{
    double value = 12.79;
    typedef double& ref_to_double;
    ref_to_double ref = value;        // Equivalent to double &ref = value;
    ref_to_double &ref2 = value;      
    ref2 = 12.111;
    cout << value << endl;        
}

输出：

12.111

后续问题：将第 8 行替换为 ref_to_double &&ref2 = value; 也不会产生编译时错误。为什么会这样？

Answer 1

Why is it possible to define a reference to a reference using type alias?

is also not producing a compile-time error. Why is this so?

why is the 8th line of this code not throwing an error?

因为语言允许这样的表达式有效，所以没有其他解释......让我们从标准中拿一个很好的例子，来自 c++draft/Declarations/References:

int i;
typedef int& LRI;
typedef int&& RRI;

LRI& r1 = i;                    // r1 has the type int&
const LRI& r2 = i;              // r2 has the type int&
const LRI&& r3 = i;             // r3 has the type int&

RRI& r4 = i;                    // r4 has the type int&
RRI&& r5 = 5;                   // r5 has the type int&&

decltype(r2)& r6 = i;           // r6 has the type int&
decltype(r2)&& r7 = i;          // r7 has the type int&