如何在C ++中从int到对象的赋值是可能的？ [英] How can the assignment from int to object be possible in C++?

问题描述

class phone {  
    public:  
        phone(int x) { num = x; }
        int number(void) { return num; }
        void number(int x) { num = x; }

    private:
        int num;
};

int main(void)
{
    phone p1(10);

    p1 = 20;    // here!

    return 0;
}

您好，伙计

Hi, guys

只是我声明了一个简单的类，如上面的一个。

之后，我分配int值到该类的对象，然后它工作！

（我打印了它的值。正确存储）

Just I declared a simple class like above one.
After that I assigned int value to the object that class, then it worked!
(I printed its value. It was stored properly)

如果没有带int参数的构造，则会发生编译错误。

所以，我认为它与构造函数有关。对不对？

If there is not a construct with int parameter, a compile error occurred.
So, I think it's related with a constructor. Is that right?

请给我一个很好的解释。

谢谢。

Please give me a good explanation.
Thanks.

推荐答案

这是合法的，因为C ++解释任何可以用 T 类型的单个参数调用的构造函数作为隐式转换 T 到自定义对象类型。在您的情况下，代码

This is legal because C++ interprets any constructor that can be called with a single argument of type T as a means of implicitly converting from Ts to the custom object type. In your case, the code

p1 = 20;

解释为

p1.operator= (20);

这又解释为

p1.operator= (phone(20));

这种行为真的很奇怪，几乎肯定不是你想要的。要禁用它，可以标记构造函数 explicit 以禁用隐式转换：

This behavior is really weird, and it's almost certainly not what you wanted. To disable it, you can mark the constructor explicit to disable the implicit conversion:

class phone {  
    public:  
        explicit phone(int x) { num = x; }
        int number(void) { return num; }
        void number(int x) { num = x; }

    private:
        int num;
};

现在，在进行隐式转换时，构造函数不会被考虑，错误。

Now, the constructor won't be considered when doing implicit conversions, and the above code will cause an error.

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