为什么允许通过const引用而不是普通引用传递R值?

问题描述：

标题说明为什么允许通过常量引用而不是常规引用传递R值(文字)

as the title says why is it allowed to pass R-Values(literals) by constant reference but not normal reference

void display(const int& a)
{
cout << a ;
}

如果被称为display(5)将起作用，但是如果没有const它将不起作用******我的意思是const引用如何保持指向R值(匿名变量)的方式*** ***

will work if called display(5) but without the const it won't work ****** I mean how can a const reference keep pointing to an R-Value (anonymous variable) ******

答

最后一个问题:

常量引用如何保持指向R值(匿名变量)

how can a const reference keep pointing to an R-Value (anonymous variable)

这是答案. C ++语言表示，本地const引用可以将临时值的生存期延长到包含作用域的末尾，但可以节省复制构造的成本(即，如果您要使用本地变量代替).

Here is the answer. The C++ language says that a local const reference prolongs the lifetime of temporary values until the end of the containing scope, but saving you the cost of a copy-construction (i.e. if you were to use an local variable instead).

为什么允许通过const引用而不是普通引用传递R值?

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