Prev: enum operator overloading VC9 compiler bug?
Next: Rectangular objects from client area
From: David Wilkinson on 1 Feb 2010 09:59
Igor Tandetnik wrote:
>> UINT_PTR n = 42;
>> LPVOID p = reinterpret_cast<LPVOID>(n);
>>
>> Many thanks. This has fixed the problem though I'm not sure why I can't just
>> do LPVOID(n)?
>
> You can. Personally, I prefer the notation using type conversion operator, but a functional cast ( LPVOID(n) ) or a C-style cast ( (LPVOID)n ) should both work, too.

Igor:

Ah, yes. I was thinking of

LPVOID p(n);

which does not work.

But I guess I am confused. I have always thought of what you call "functional
cast" as invoking the constructor. So why exactly do

LPVOID p = LPVOID(n);
LPVOID p(LPVOID(n));

both work, while

LPVOID p = n;
LPVOID p(n);

do not?

--
David Wilkinson
Visual C++ MVP
From: Bo Persson on 1 Feb 2010 11:44
David Wilkinson wrote:
> Igor Tandetnik wrote:
>>> UINT_PTR n = 42;
>>> LPVOID p = reinterpret_cast<LPVOID>(n);
>>>
>>> Many thanks. This has fixed the problem though I'm not sure why I
>>> can't just do LPVOID(n)?
>>
>> You can. Personally, I prefer the notation using type conversion
>> operator, but a functional cast ( LPVOID(n) ) or a C-style cast
>> ( (LPVOID)n ) should both work, too.
>
> Igor:
>
> Ah, yes. I was thinking of
>
> LPVOID p(n);
>
> which does not work.
>
> But I guess I am confused. I have always thought of what you call
> "functional cast" as invoking the constructor.

It can invoke a constructor, it the type has one. For built-in types
it is the same as a C-style cast.

> So why exactly do
>
> LPVOID p = LPVOID(n);
> LPVOID p(LPVOID(n));

Because these are casts.

>
> both work, while
>
> LPVOID p = n;
> LPVOID p(n);
>
> do not?

And these are not. LPVOID doesn't have a converting constructor.


Bo Persson




From: Igor Tandetnik on 1 Feb 2010 11:41
David Wilkinson <no-reply(a)effisols.com> wrote:
> But I guess I am confused. I have always thought of what you call
> "functional cast" as invoking the constructor.

When a functional cast expression has one parameter, it is exactly equivalent to a corresponding C-style cast. Yes, you can invoke a constructor using C-style cast:

class C {
public:
C(int);
};

C v = (C)42;

> So why exactly do
>
> LPVOID p = LPVOID(n);
> LPVOID p(LPVOID(n));
>
> both work, while
>
> LPVOID p = n;
> LPVOID p(n);
>
> do not?

Well, because the standard says so. I don't know the rationale behind this, but if I had to guess, I'd say the committee made functional cast and C-style cast behave the same in order to avoid the proliferation of similar but subtly different cast mechanisms.
--
With best wishes,
Igor Tandetnik

With sufficient thrust, pigs fly just fine. However, this is not necessarily a good idea. It is hard to be sure where they are going to land, and it could be dangerous sitting under them as they fly overhead. -- RFC 1925

First  |  Prev  | 
Pages: 1 2
Prev: enum operator overloading VC9 compiler bug?
Next: Rectangular objects from client area