Const and non-const accesors, which will be called

Asked Sep 24 '18 at 13:51

Active Sep 24 '18 at 13:51

Viewed 59 times

I have a simple question but couldnt find the answer : what is the rule that the standard follows to choose between a const and non-const function.

Here is an example

#include <iostream>
class B{
int a;

public:

B(int _a) : a(_a){}

const int& getA() const{
    std::cout << "CONST\n";
    return a;
}

int& getA(){
    std::cout << "NON CONST\n";
    return a;
}

};

int main(){

B b(3);

b.getA(); //print non const
int a = 3 + b.getA(); //same
};

Both example use the non-const version. I thought it would pick the const version, since no modification is done.

asked Sep 24 '18 at 13:51

Armassarion

1

Try with a `const` object of type `B`... – Adriano Repetti Sep 24 '18 at 13:53
"_I thought it would pick the const version, since no modification is done._" `const` variant is only picked, when invoked on a `const` instance. What underlying implementation does is irrelevant. – Algirdas Preidžius Sep 24 '18 at 13:53
2

How can you tell that no modification is done *before* you choose an overload? – molbdnilo Sep 24 '18 at 13:56

Const and non-const accesors, which will be called

0 Answers0