为什么一个const成员函数可以修改静态数据成员？ [英] Why can a const member function modify a static data member?

问题描述

在下面的 C ++ 程序中，修改静态数据成员从 const 函数工作正常：

  class A 
 {
 public：
 static int a; //静态数据成员
 
 void set（）const 
 {
 a = 10; 
} 
}; 
  

但是从<$ c $修改非静态数据成员 c> const 函数不起作用：

  class A 
 {
 public：
 int a; //非静态数据成员
 
 void set（）const 
 {
 a = 10; 
} 
}; 
  

为什么 const 成员函数修改一个静态数据成员？

解决方案

规则，就是这样。
$ b $

成员函数上的 const 限定符意味着您不能修改非 - mutable 非 - 静态类成员变量。

一些合理化，在 const 限定成员函数中 this 指针是一个 const 类型，并且这个与类的一个实例有内在联系。 static 成员与类实例无关。你不需要一个实例来修改 static 成员：你可以通过编写 A :: a = 10; 。

因此，在第一种情况下，请将 a = 10; 看作简写为 A :: a = 10; ，在第二种情况下，将其视为 this-> a = 10的简写形式; code>，由于 this 的类型是 const A * 。

In the following C++ program, modifying a static data member from a const function is working fine:

class A 
{
  public:   
    static int a; // static data member

    void set() const
    {
        a = 10;
    }
};

But modifying a non-static data member from a const function does not work:

class A 
{
  public:   
    int a; // non-static data member

    void set() const
    {
        a = 10;
    }
};

Why can a const member function modify a static data member?

解决方案

It's the rule, that's all. And for good reason.

The const qualifier on a member function means that you cannot modify non-mutable non-static class member variables.

By way of offering some rationalisation, the this pointer in a const qualified member function is a const type, and this is inherently related to an instance of a class. static members are not related to a class instance. You don't need an instance to modify a static member: you can do it, in your case, by writing A::a = 10;.

So, in your first case, think of a = 10; as shorthand for A::a = 10; and in the second case, think of it as shorthand for this->a = 10;, which is not compilable since the type of this is const A*.

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