使用静态类型转换时，将其转换为类类型或对象引用是否更有意义？

Question

I just realized that I can use either to achieve the same effect. Are there any caveats? Which convention makes more sense?

#include <iostream>

using namespace std;

class some_class {
public:
    void echo() {
        cout << 1 << endl;
    }
};

class other_class : public some_class {
public:
    void echo() {
        // Does it matter? 
        static_cast<some_class>(*this).echo();
        static_cast<some_class&>(*this).echo();
    }
};

int main() {
    other_class a;

    a.echo();

    return 0;
}

Answer 1

The first cast creates a temporary object of type some_class , initialized from *this by slicing, then calls echo on the temporary object, then destroys the temporary object. If you made the echo function update a member variable of some_class , you would notice that *this actually did not get updated.

The second cast calls some_class::echo(); function on the *this object, without creating anything.

Typically the second option is what you are aiming for. As noted by davidhigh in comments, it is a cleaner code style (IMHO anyway) to simply write some_class::echo(); instead of using the cast.