How to directly call conversion operator?

Question

struct Bob
{
    template<class T>
    void operator () () const
    {
        T t;
    }

    template<class T>
    operator T () const
    {
        T t;
        return t;
    }
};

I can directly call Bob's operator() like this

Bob b;
b.operator()<int>();

How to directly call the conversion operator with a specific template parameter like this?

Bob b;
std::string s = b.???<std::string>();

It's not possible to use static_cast

Bob b;
std::string s = static_cast<std::string>(b);

http://ideone.com/FoBKp7

error: call of overloaded ‘basic_string(Bob&)’ is ambiguous

Question How to call directly with template parameter OR it's not possible. I know there are workarounds using a wrapping function.

hah, nice try. No, not possible as far as I'm aware. that's why I suggested using a real function. Besides, what's the point of making it a conversion function if you're going to call it like a regular one? — Dave, Aug 30 '13 at 18:19
@Dave The same for the other operators. A backup in case you want to for some reason. — Neil Kirk, Aug 30 '13 at 18:20
Or even `b.operator decltype(s)();` (which I was admittedly surprised worked in clang). Then again, so does simple `std::string s = b;` so I guess that makes sense. — WhozCraig, Aug 30 '13 at 18:24
Well there you are. Now I've learned two things about cast functions today. (I'll likely never use either because they're ugly but hey, knowledge) — Dave, Aug 30 '13 at 18:24

score 20 · Accepted Answer · edited May 23 '17 at 12:09

20

You can call it directly (explicitly) like this:

Bob b;
std::string s = b.operator std::string();

but it's not "with a specific template parameter" (but there's no need for).

See also WhozCraig's comment

edited May 23 '17 at 12:09

Community

1
1

answered Aug 30 '13 at 18:27

gx_

4,690
24
31

1

Fwiw, `std::string s = b;` also works (at least on my rig). But thats beginning to paddle away from the original question at a pretty rapid rate =) – WhozCraig Aug 30 '13 at 18:29
@WhozCraig Of course (the purpose of a conversion operator is to enable implicit conversion after all). But I linked to your other comment that adds `decltype(s)` :) – gx_ Aug 30 '13 at 18:31

score 3 · Answer 2 · answered Aug 30 '13 at 18:17

3

Use a helper function:

template< typename T >
T explicit_cast( T t ) { return t; }

int main()
{
    Bob b;
    std::string s = explicit_cast<std::string>(b);
}

answered Aug 30 '13 at 18:17

Daniel Frey

55,810
13
122
180

4

*cough*... value parameter ... *cough*. – WhozCraig Aug 30 '13 at 18:19
I want to know if it is possible or not to call directly. – Neil Kirk Aug 30 '13 at 18:19
@WhozCraig I think in this case a by-value parameter is better than a by-const-reference as it allows more optimization. Or do you think otherwise? – Daniel Frey Aug 30 '13 at 18:23
I think *both* value and reference should be supplied via overload, but then again, I'm not sure I'd ever find myself in a position to do *either* =) – WhozCraig Aug 30 '13 at 18:25
1

Your "explicit_cast" resembles [`implicit_cast`](http://stackoverflow.com/a/863761) (which is better defined) – gx_ Aug 30 '13 at 18:40

How to directly call conversion operator?

2 Answers2

Linked