如何重载的const和非const函数工作？ [英] how does overloading of const and non-const functions work?

问题描述

stl充满了这样的定义：

The stl is full of definitions like this:

iterator begin ();
const_iterator begin () const;

由于返回值不参与重载解析，唯一的区别是函数 const 。这是过载机制的这一部分吗？编译器的算法用于解析一行，如：

As return value does not participate in overloading resolution, the only difference here is the function being const. Is this part of overloading mechanism? What is the compiler's algorithm for resolving a line like:

vector<int>::const_iterator it = myvector.begin();

推荐答案

在您给出的示例中：

vector<int>::const_iterator it = myvector.begin();

如果 myvector -const版本 begin（）将被调用，你将依赖于从迭代器到const_iterator的隐式转换。

if myvector isn't const the non-const version of begin() will be called and you will be relying on an implicit conversion from iterator to const_iterator.

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