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Date: Tue, 08 Jan 2002 11:57:21 +0100
To: "William A. Rowe, Jr." <wrowe@covalent.net>,dev@apr.apache.org
From: Christian Gross <christiangross@yahoo.de>
Subject: Re: Problem with pointer conversion
Cc: <dev@apr.apache.org>
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At 20:52 07/01/2002 -0600, William A. Rowe, Jr. wrote:
>From: "Branko =C8ibej" <brane@xbc.nu>
>Sent: Monday, January 07, 2002 8:29 PM
>
>
> > A void pointer is implicitly converted to any other pointer type in C,
> > so this cast shouldn't be necessary. Can you be a bit more specific
> > about the compiler you're using, the platform, and the error message you
> > get?
> >

This is when I compile the code using C++.  In C++ there is no explicit=20
type conversion.  The problem is that the macro is included in a header,=20
which is then translated as C++.


>No.  It breaks type saftey.

This is a C++ feature.


>Because you've already created the terrific opportunity to segfault.
>This bug on Win32 is usually caused by declaring something without your
>own FOO_DECLARE or FOO_DECLARE_NONSTD wrappers (or the wrong case.)

Already ran into that, but it caused different compile error.  Similar, but=
=20
not the same.  Consider the following code, which is a simplified version=20
of the macro:

void *SomeVoidFunc() {
         void *ptr =3D NULL;
         return ptr;
}

void AnotherFunc() {
         int *myval;

         myval =3D SomeVoidFunc();  // In C this is legal, but C++ causes a=
=20
compile fault
         myval =3D (int *)SomeVoidFunc();  // For C++ to compile need a=20
pointer conversion
}

The type cast performed in the macro is identical to the value of=20
pointer.  So if the macro was properly declared then the typecast will be=20
properly performed.

Christian