为什么要使用const和非const访问器？

问题描述

为什么STL容器定义访问器的const和非const版本？

Why do the STL containers define const and non-const versions of accessors ?

定义 const T&有什么好处？ at（unsigned int i）const 和 T& at（unsigned int）和非const版本？

What is the advantage of defining const T& at(unsigned int i) const and T& at(unsigned int) and not only the non-const version ?

推荐答案

因为您不会不能在 const 矢量对象上调用 at 。

Because you wouldn't be able to call at on a const vector object.

如果您只有非 const 版本，则以下内容：

If you only had the non-const version, the following:

const std::vector<int> x(10);
x.at(0);

不会编译。拥有 const 版本可以实现这一点，同时也可以防止您实际更改 at 返回的内容-根据合同，因为向量是 const 。

would not compile. Having the const version makes this possible, and at the same time prevents you from actually changing what at returns - which is by contract, since the vector is const.

非 const const 对象上调用c $ c>版本，并允许您修改返回的元素，这也是有效的，因为矢量不是const。

The non-const version can be called on a non-const object and allows you to modify the returned element, which is also valid because the vector isn't const.

const std::vector<int> x(10);
      std::vector<int> y(10);

int z = x.at(0);          //calls const version - is valid
x.at(0) = 10;             //calls const version, returns const reference, invalid

z = y.at(0);              //calls non-const version - is valid
y.at(0) = 10;             //calls non-const version, returns non-const reference
                          //is valid

这篇关于为什么要使用const和非const访问器？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！