Const引用和左值 [英] Const reference and lvalue

问题描述

我们不能写 int& ref = 40 ，因为我们需要在右侧的 lvalue 。但我们可以写 const int& ref = 40 。为什么会这样？ 40是 rvalue 而不是 lvalue

const T&不需要是左值或者甚至类型
T。在这种情况下：

[1]首先，如果必要，应用到T的隐式类型转换。 / p>

[2]然后，将结果值放在
类型T的临时变量中。

[3]最后，这个临时变量用作
初始化程序的值。

const int& ref = 40 ，临时int变量在幕后创建，ref绑定到这个临时变量。

We cannot write int& ref = 40 because we need lvalue on right side. But we can write const int& ref = 40 . Why is this possible? 40 is rvalue instead lvalue

I know that this is an exception but why?

解决方案

As Stroustrup says:

The initializer for a const T& need not be an lvalue or even of type T. In such cases:

[1] First, implicit type conversion to T is applied if necessary.

[2] Then, the resulting value is placed in a temporary variable of type T.

[3] Finally, this temporary variable is used as the value of the initializer.

So, when you type const int& ref = 40, the temporary int variable is created behind the scenes, and ref is bound to this temporary variable.

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