Hello,

I have two operators, one returning a value

int operator()(int i) const

and second allowing to get the reference of a value (to write it):

int& operator()(int i)

And the problem is that no matter what always the second operator is called. Shouldn't it be called only when reference is needed?

I have written a simple class illustrating my problem:

#include <iostream>

class Test
{
private:
    int tab[10],N;

public:
    Test()
    {
      N=10;
      for (int i=0;i<N;i++) tab[i]=i;
    }

    int operator()(int i) const
    {
      std::cout<<"why am i never called?"<<std::endl;
      return tab[i];
    }

    int& operator()(int i)
    {
      std::cout<<"why am i always called?"<<std::endl;
      return tab[i];
    }
};

And the following code demonstrates the issue:

	Test a;
	int b=a(1);  //uses the &operator
        a(2); //uses the &operator
        a(3)=8; //uses the &operator

I'll be grateful for any suggestions.

The const-version of the operator would only be called on a const instance.

The const-version of the operator would only be called on a const instance.

But when there is no

int& operator()(int i)

then the

int operator()(int i) const

is called. Of course then I can only read an element. Is there any way to use both of those operators, dependently on the context?

Thank you for your reply.

Why is this a problem?

$a = 6, 1, 1, 1, 7; say $a; # 6
$a = (6, 1, 1, 1, 7); say $a; # 7
@a = (6, 1, 1, 1, 7); $a = @a; say $a; # 5

By default C++ utilizes operator overloading in operation chaining. Thats why it prefers the method, returning l-value. Temporal objects cannot be l-values.