C ++类型转换运算符

Question

I am studying operator overloading, there are some parts that are difficult to understand.

See this example code.

class A {
    private:
    char a;
    int b;
    double c;

public:
A(char _a = 'a', int _b = 99, double _c = 1.618) :a(_a), b(_b), c(_c){
}

public:
    operator char() const {
        cout << "operator char() called" << endl;
        return this->a;
    }

operator int() const {
        cout << "operator int() called" << endl;
        return this->b;
    }

operator double() {
        cout << "operator double() called" << endl;
        return this->c;
    }
};
int main(void) {
    A a;
    char b = a;
    int c = a;
    double d = a;

    printf("%c\n", b);
    printf("%d\n", c);
    printf("%f\n", d);

    return 0;
}

I made this code to test for type conversion operator and expected that the appropriate function would be called for each type of data.

But the result is..

operator double() called
operator double() called
operator double() called
    <-- strange character is gone on board!
1
1.618000

I can not understand why the results are not as follows.

operator char() called
operator int() called
operator double() called
a
99
1.618

Why is double operator called when converting to char and int?

Have a good day! :)

Answer 1

You forgot the const on the double conversion operator:

operator double() const {  // <---------------------------
        cout << "operator double() called" << endl;
        return this->c;
    }
};

As in your example a is not const , the double conversion is the best match. If you fix that you get the expected output.

Live example

...some opinion based PS:

I didnt find what the core guidelines say about conversion operators, but if I had to make up a guideline for conversion operators it would be: Avoid them. If you use them, make them explicit . The surprising effects of implicit conversion outweigh the benefits by far.

Just as an example, consider std::bitset . Instead of offering conversion operators it has to_string , to_ulong and to_ullong . It is better to have your code explicit. A a; double d = a; is a little bit mysterious. I would have to look at the class definition to get an idea of what is really going on. On the other hand A a; double d = a.as_double(); A a; double d = a.as_double(); can do the exact same thing, but is way more expressive.

Answer 2

Yea so the problem is, that you made all operators const except for the double operator. I am still a bit surprised because this const just means that the operator call does not modify the class members. Still it seems that only the double operator is called for all 3. I would all 3 op's make const and then it will work properly.

If someone has an explanation why this happens, I would also like to know. Cheers.

operator char() const { // here is const
    cout << "operator char() called" << endl;
    return this->a;
}

operator int() const { // here is const
    cout << "operator int() called" << endl;
    return this->b;
}

operator double() { // here is no const
    cout << "operator double() called" << endl;
    return this->c;
}