C++重载转换运算符

Question

我试图有一个允许隐式转换为某些内置类型的类,比如unsigned long int,因为我试图尽可能正确地做这个(这是我在C++中的第一个重要项目),我遇到了一个奇怪的关于const正确性的问题:

这有效:

#include <iostream>

class CustomizedInt
{
private:
    int data;
public:
    CustomizedInt();
    CustomizedInt(int input);
    operator unsigned long int () const
    {
        unsigned long int output;
        output = (unsigned long int)data;
        return output;
    }
};

CustomizedInt::CustomizedInt()
{
    this->data = 0;
}

CustomizedInt::CustomizedInt(int input)
{
    this->data = input;
}

int main()
{
    CustomizedInt x;
    unsigned long int y = x;

    std::cout << y << std::endl;

    return 0;
}

但是这个:

#include <iostream>

class CustomizedInt
{
private:
    int data;
public:
    CustomizedInt();
    CustomizedInt(int input);
    operator unsigned long int () const;
};

CustomizedInt::CustomizedInt()
{
    this->data = 0;
}

CustomizedInt::CustomizedInt(int input)
{
    this->data = input;
}

CustomizedInt::operator unsigned long()
{
    unsigned long int output;
    output = (unsigned long int)data;
    return output;
}

int main()
{
    CustomizedInt x;
    unsigned long int y = x;

    std::cout << y << std::endl;

    return 0;
}

在Visual Studio 2010中给我这个错误: error C2511: 'CustomizedInt::operator unsigned long(void)' : overloaded member function not found in 'CustomizedInt'

现在,如果我从运算符定义中删除关键字const,一切正常.这是一个错误吗？我读到我应该在每个(公共)方法/运算符之后使用const关键字,以便清楚地声明它不会以任何方式改变当前对象.

此外,我知道定义这样的运算符可能是不好的做法,但我不确定我是否完全理解相关的警告.有人可以概述一下吗？仅仅定义一个名为ToUnsignedLongInt的公共方法会更好吗？

Answer 1

函数签名与函数定义不匹配.

operator unsigned long int () const;

和

CustomizedInt::operator unsigned long()    { ... }
                                       ^^^
                                   const missing

在这种情况下,您应该将转换运算符标记为const因为它不会影响对象的内部状态.

此外,使用构造函数初始化列表初始化成员变量.

CustomizedInt::CustomizedInt()
: data()
{
}

CustomizedInt::CustomizedInt(int input)
: data(input)
{
}

Answer 2

您可以const从声明中删除它,但您几乎肯定想要做的是将其添加到定义中:

CustomizedInt::operator unsigned long() const
{
    unsigned long int output;
    output = (unsigned long int)data;
    return output;
}