Does C++ automatically commute some operation overloads?

user_15282553 0
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I've been trying to understand operation overloading and the commutative property between classes and fundamental types. I declared the Test to Int asterisk operation but not the Int to Test, just to see if C++ would automatically provide with some sort of commutative function. The testing code is:

#include <iostream>

class Test{
public:
    float x;

    Test(){
        x = 1;
    }
    Test(float number){
        x = number;
    }
    ~Test(){}
};

Test operator*(Test a, int b){
    std::cout << "Alert! Test to Int happened! ";
    return Test(a.x * b);
}

Test operator*(Test a, Test b){
    std::cout << "Alert! Test to Test happened! ";
    return Test(a.x * b.x);
}

int main(){
    
    Test my_test;
    std::cout << "First result: " << (my_test * 2).x << std::endl;
    std::cout << "Second result: " << (3 * my_test).x << std::endl;
    
    return 0;
}

with output:

First result: Alert! Test to Int happened! 2
Second result: Alert! Test to Test happened! 3

The first result was as expected, but for the second result I expected either an error or some unknown-to-me function to do the work. Instead, the Test to Test operation was used... Can someone explain why and what happened?

operator-overloading c++
1 Answers

Your constructor with the form of

Test(float number){
    x = number;
}

is what is called a converting constructor. Since it is not explicit, it allows you convert a float to a Test implicitly. You see that in

(3 * my_test).x

Here the 3 gets converted to a float in the standard conversion sequence and then that float get converted via the Test(float number) constructor. This then allows operator*(Test a, Test b) to be used.

To stop this, you make the constructor explicit like

explicit Test(float number){
    x = number;
}
user_43424980
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