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Does C++ provide a means to use a conversion operator as a lval?

ie, I want:

class superInt
{
public:
operator int() { return this->value; }
private:
int value;
};

....to instead be this:

class superInt
{
public:
int& operator int() { return this->value; }
private:
int value;
};

....which will not compile.

I suspect the C++ language designers avoided this on purpose, but I
thought I would ask here just in case there is a means to do this (in
some other way then I present above).

Such a thing would save me time re-writing a bunch of overloaded
operators for superInt (among other things).

-Matt
--
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Matt wrote:
> Does C++ provide a means to use a conversion operator as a lval?\

Sure.

> ie, I want:
>
> class superInt
> {
> public:
> operator int() { return this->value; }

Should probably be

operator int() const { return this->value; }

> private:
> int value;
> };
>
> ...to instead be this:
>
> class superInt
> {
> public:
> int& operator int() { return this->value; }

operator int&() { return this->value; }

> private:
> int value;
> };
>
> ...which will not compile.

Of course.

>
> I suspect the C++ language designers avoided this on purpose,

No, they didn't.

> but I
> thought I would ask here just in case there is a means to do this (in
> some other way then I present above).

Very good choice.

> Such a thing would save me time re-writing a bunch of overloaded
> operators for superInt (among other things).

But it is rather dangerous. Be careful.

Victor

On Mon, 31 Jan 2005 12:01:17 -0500, Victor Bazarov
<> wrote:
>
> operator int&() { return this->value; }

Great, I'm glad I asked here. Thanks!

>
>> Such a thing would save me time re-writing a bunch of overloaded
>> operators for superInt (among other things).
>
>But it is rather dangerous. Be careful.

Yes, I can see where it can be quite dangerous. I think I will avoid
it whenever I can...especially since I've already written all my
overloaded operators for the task at hand.

-Matt
--
Remove the "downwithspammers-" text to email me.