Why having const and non-const accessors?(为什么要有 const 和非 const 访问器?)
问题描述
Why do the STL containers define const and non-const versions of accessors ?
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 ?
Because you wouldn't be able to call at
on a const
vector object.
If you only had the non-const
version, the following:
const std::vector<int> x(10);
x.at(0);
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
.
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
这篇关于为什么要有 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编程学习网!
本文标题为:为什么要有 const 和非 const 访问器?


基础教程推荐
- 如果我为无符号变量分配负值会发生什么? 2022-01-01
- 为什么派生模板类不能访问基模板类的标识符? 2021-01-01
- 非静态 const 成员,不能使用默认赋值运算符 2022-10-09
- 为什么 RegOpenKeyEx() 在 Vista 64 位上返回错误代码 2021-01-01
- 通过引用传递 C++ 迭代器有什么问题? 2022-01-01
- CString 到 char* 2021-01-01
- 为什么 typeid.name() 使用 GCC 返回奇怪的字符以及如 2022-09-16
- 我应该对 C++ 中的成员变量和函数参数使用相同的名称吗? 2021-01-01
- GDB 显示调用堆栈上函数地址的当前编译二进制文 2022-09-05
- 初始化列表*参数*评估顺序 2021-01-01