函数只接受非常量左值(Having a function only accept non-const lvalues)

 我有一个函数，它将第一个向量排序为两个向量作为排序条件。 它的签名是 
 
template<typename A, typename B>
void sort(A&& X, B&& Y)
{
  ..
}
 
 问题是通用引用会允许像这样的无意义的情况 
 
sort(vector<int>{ 2,1,3 }, vector<int>{ 3,1,2 });
 
 之后右值将被销毁（废话）。 
 
 明确询问左值是不起作用的 
 
template<typename A, typename B>
void sort(A& X, B& Y) ... // (*)

sort(vector<int>{2,1,3}, vector<int>{3,1,2});
 
 由于某种原因上面的编译（我认为只允许const值左右绑定到rvalues并延长它们的生命周期？）。 
 
 如果我将const添加到左值引用，那么函数将不再能够修改矢量并对它们进行排序。 
 
 
 我的问题是： 
 
 1）为什么在标有// (*)的示例中，我可以将rvalue绑定到一个甚至不是const的左值？ 为什么要像int& r = 20; 不允许？ 有什么不同？ 
 
 2）我该如何解决我的问题，即让函数只接受左值而不是临时值？ （如果可能的话，当然） 
 
 很明显，我可以使用任何可用的C ++版本 

I have a function which sorts two vectors with the first of them as ordering criterion. Its signature is
 
template<typename A, typename B>
void sort(A&& X, B&& Y)
{
  ..
}
 
The problem is that universal references would allow nonsense cases like
 
sort(vector<int>{ 2,1,3 }, vector<int>{ 3,1,2 });
 
where an rvalue will be destroyed afterwards (nonsense).
 
Asking explicitly for a lvalue doesn't work since
 
template<typename A, typename B>
void sort(A& X, B& Y) ... // (*)

sort(vector<int>{2,1,3}, vector<int>{3,1,2});
 
for some reason the above compiles (I thought only const lvalues were allowed to bind to rvalues and to prolong their lifetime?). 
 
If I add const to the lvalue reference then the function will no longer be able to modify the vectors and sort them.
 
 
My questions are:
 
1) Why in the example marked with // (*) can I bind a rvalue to a lvalue that is not even const ? Why instead something like int& r = 20; isn't allowed? What's the difference?
 
2) How can I solve my issue i.e. having the function accept only lvalues and not rvalue temporaries? (If it's possible, of course)
 
Obviously I'm allowed to use any C++ version available

原文：https://stackoverflow.com/questions/32720767