了解引用绑定

Question

我们不能将非 const 左值引用绑定到右值，但它可以绑定到 const 引用。我们也不能将右值引用绑定到左值。实际上标准是这样说的：

8.5.3/5.2:

the reference shall be an lvalue reference to a non-volatile const type (i.e., cv1 shall be const), or the reference shall be an rvalue reference.

但是有比"The Standard say so"更好的解释吗？

Answer 1

因为它没有语义意义。

您不能将非常量左值引用绑定到右值，因为修改右值意味着什么？根据定义，没有其他人会看到结果，所以这没有意义。

int& i = 3;
//should change referenced location, but we aren't referencing a memory location
i = 5;

您不能将右值引用绑定到左值，因为右值引用的存在是为了促进对其引用的破坏性优化。你不希望你的物体被任意从你下面移出，所以标准不允许它。

void process_string (std::string&&);
std::string foo = "foo";
//foo could be made garbage without us knowing about it
process_string (foo); 
//this is fine
process_string (std::move(foo));

Answer 2

想一些真实的案例：

#include <vector>

void f1( int& i ){i = 1;}
void f2( const int& i ){i = 1;}
void f3( std::vector<int>&& v ){auto v_move{v};}

int main()
{
    f1(3); // error: how can you set the value of "3" to "1"?
    f2(3); // ok, the compiler extend the life of the rvalue "into" f2
    std::vector<int> v{10};
    f3(v); // error: an innocent looking call to f3 would let your v very different from what you would imagine
    f3(std::vector<int>{10}); // ok, nobody cares if the rvalue passed as an argument get modified
}

了解引用绑定

Understanding reference binding

c++

reference