为什么 STL 映射的 [] 运算符不是 const? [英] Why isn't the [] operator const for STL maps?

问题描述

为了这个问题，人为的例子:

Contrived example, for the sake of the question:

void MyClass::MyFunction( int x ) const
{
  std::cout << m_map[x] << std::endl
}

这不会编译，因为 [] 运算符是非常量的.

This won't compile, since the [] operator is non-const.

这很不幸，因为 [] 语法看起来很干净.相反，我必须这样做:

This is unfortunate, since the [] syntax looks very clean. Instead, I have to do something like this:

void MyClass::MyFunction( int x ) const
{
  MyMap iter = m_map.find(x);
  std::cout << iter->second << std::endl
}

这一直困扰着我.为什么 [] 运算符是非常量的?

This has always bugged me. Why is the [] operator non-const?

推荐答案

对于std::map和std::unordered_map，operator[] 会将索引值插入到容器中.这有点不直观，但就是这样.

For std::map and std::unordered_map, operator[] will insert the index value into the container if it didn't previously exist. It's a little unintuitive, but that's the way it is.

由于必须允许失败并插入默认值，因此该运算符不能用于容器的 const 实例.

Since it must be allowed to fail and insert a default value, the operator can't be used on a const instance of the container.

http://en.cppreference.com/w/cpp/container/map/operator_at

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