Why a function with const type reference argument doesn't use it as a rvalue ?

[edlep]

Hello,

Given a class with [] operators for both lvalue and rvalue:

class A
{
  public:
    double items[10];
    double operator [] (unsigned int i)
    {
        cout << "rvalue operator" << endl;
        return items[i];
    }
    double &operator [] (unsigned int i) const
    {
        cout << "lvalue operator" << endl;
        return items[i];
    }
};

and a function with a const A& as argument

void theFunc(const A &a)
{
}

The following lines have the expected behavior:

A a;
double v = a[0]; // calls the rvalue [] operator
a[1] = 10; // calls the lvalue [] operator

But I wouldn't have expected this one:

theFunc(a); // calls the rvalue operator ! Why ?

Obviously, if I cast the argument with something like this

theFunc(static_cast<const A&>(a));

, it works as expected, but I'd rather use a cleaner solution.

Why the rvalue isn't used in that case ?
Does anyone know a solution to this issue ?

Thanks.

Ed

[grumpy]

With the code snippets you have shown, calling theFunc(a) cannot call either of the [] operators. It simply passes a const reference to a.

If you're getting some other result, it probably means theFunc() is overloaded (i.e. there is a version which takes some type of argument, other than what you have shown).

[edlep]

[EDIT]
You are absolutely right, sorry for that error.
The idea was obviously the following:

void theFunc(const double &d)
{
}

and the difference between

theFunc(a[0]); // which uses the lvalue

// and

theFunc(static_cast<const double&>(a[0])); // which uses the rvalue

[grumpy]

The compiler avoids creating temporaries if possible, in order to pass an argument by reference. The operator that returns a reference is therefore a better match than one which returns by value.

The static_cast<> forces the compiler to generate a const reference to the value returned by your "rvalue" form.

A better solution would be to pass theFunc()'s argument by value.

[edlep]

Grumpy,
Thanks for your answer.
What do you mean by

pass theFunc()'s argument by value.

?


By the way, I realize that I also made ​​a mistake in the definition of my example class (the const is on the wrong operator) ... I should sleep a bit before posting.
The right definition is:

class A
{
  public:
    double items[10];
    double operator [] (unsigned int i) const
    {
        cout << "rvalue operator" << endl;
        return items[i];
    }
    double &operator [] (unsigned int i)
    {
        cout << "lvalue operator" << endl;
        return items[i];
    }
};

Sorry for the confusion of my post.

[tabstop]

Grumpy,
Thanks for your answer.
What do you mean by pass theFunc()'s argument by value ?

You are currently passing the argument to theFunc by reference:

void theFunc(const double &d)