类的函数声明中最后一个“const"的含义?

2022-01-23 00:00:00 constants c++ declaration c++-faq

const 在这样的声明中是什么意思?const 让我很困惑.

What is the meaning of const in declarations like these? The const confuses me.

class foobar
{
  public:
     operator int () const;
     const char* foo() const;
};

推荐答案

当你将 const 关键字添加到方法时，this 指针本质上将变成指向 const 对象，因此您不能更改任何成员数据.(除非你使用 mutable，稍后会详细介绍).

When you add the const keyword to a method the this pointer will essentially become a pointer to const object, and you cannot therefore change any member data. (Unless you use mutable, more on that later).

const 关键字是函数签名的一部分，这意味着您可以实现两种类似的方法，一种在对象为 const 时调用，另一种在对象为 const 时调用.不.

The const keyword is part of the functions signature which means that you can implement two similar methods, one which is called when the object is const, and one that isn't.

#include <iostream>

class MyClass
{
private:
    int counter;
public:
    void Foo()
    { 
        std::cout << "Foo" << std::endl;    
    }

    void Foo() const
    {
        std::cout << "Foo const" << std::endl;
    }

};

int main()
{
    MyClass cc;
    const MyClass& ccc = cc;
    cc.Foo();
    ccc.Foo();
}

这将输出

Foo
Foo const

在非常量方法中，您可以更改实例成员，而在 const 版本中则无法做到这一点.如果你把上面例子中的方法声明改成下面的代码，你会得到一些错误.

In the non-const method you can change the instance members, which you cannot do in the const version. If you change the method declaration in the above example to the code below you will get some errors.

    void Foo()
    {
        counter++; //this works
        std::cout << "Foo" << std::endl;    
    }

    void Foo() const
    {
        counter++; //this will not compile
        std::cout << "Foo const" << std::endl;
    }

这并不完全正确，因为您可以将成员标记为 mutable，然后 const 方法可以更改它.它主要用于内部计数器和东西.解决方案是下面的代码.

This is not completely true, because you can mark a member as mutable and a const method can then change it. It's mostly used for internal counters and stuff. The solution for that would be the below code.

#include <iostream>

class MyClass
{
private:
    mutable int counter;
public:

    MyClass() : counter(0) {}

    void Foo()
    {
        counter++;
        std::cout << "Foo" << std::endl;    
    }

    void Foo() const
    {
        counter++;    // This works because counter is `mutable`
        std::cout << "Foo const" << std::endl;
    }

    int GetInvocations() const
    {
        return counter;
    }
};

int main(void)
{
    MyClass cc;
    const MyClass& ccc = cc;
    cc.Foo();
    ccc.Foo();
    std::cout << "Foo has been invoked " << ccc.GetInvocations() << " times" << std::endl;
}

哪个会输出

Foo
Foo const
Foo has been invoked 2 times

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