为什么不能使用右值来初始化左值引用? [英] Why a rvalue cannot be used to initialize a lvalue reference?

问题描述

我可以做 const A&a = A()和 A&&a = A()，但是为什么我不能做 A&a = A()?

I can do const A& a = A() and A&& a = A(), but why cannot I do A& a = A()?

推荐答案

基本原理是，对临时变量进行突变几乎没有任何意义.这样做的任何尝试都可能是编程错误，而不是刻意的决定.

The rationale is that it rarely makes sense to mutate a temporary. Any attempt to do so is likely to be a programming error rather than a deliberate decision.

典型示例如下.假定允许绑定到非常量引用.

The prototypical example is as follows. Assume binding to non-const reference is allowed.

void foo(int& x, int& y); // sets x and y
int xx, yy;
foo(xx, yy); // xx and yy are set
// now make a small change...
void foo(long& x, long& y); // upgrade
int xx, yy; // forgot to change here 
foo(xx, yy); // silently breaks

尽管有时确实想要对临时变量进行突变，但这通常是出于与对左值进行突变的原因完全不同的原因.发明了右值引用以适应这些情况.

While one does sometimes want to mutate a temporary, it's often for an entirely different reason from that of mutating an lvalue. Rvalue references were invented to accommodate these cases.

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