Re: [GD-General] meaning of sizeof(int) on all plateform

[brian s. <pud...@po...>]

It is wrong to assume that all pointers are the same size, even on 
"normal" architectures.  Pointers to class members are not required to 
be sizeof(void *), and in the case of virtual inheritance, probably 
aren't. 

This program:

    // Compile under VC++ with "cl /ML /GX test.cpp"

    #include <iostream>
    using namespace std;

    struct Base
    {
    };

    struct Derived : public virtual Base
    {
    };

    typedef void (Base::*BasePtrToMember)();
    typedef void (Derived::*DerivedPtrToMember)();

    int main(int argc, char **argv)
    {
    cout << "sizeof(BasePtrToMember) " << sizeof(BasePtrToMember) << endl;
    cout << "sizeof(DerivedPtrToMember) " << sizeof(DerivedPtrToMember) 
<< endl;

    return 1;
    }

Produces the following output:

    sizeof(BasePtrToMember) 4
    sizeof(DerivedPtrToMember) 12

There you go, a 12 byte pointer on Win32.  You can imagine that this 
causes brain-bending bugs for the unwary :)

--brian

Jay Woodward wrote:

>Hold on -- it seems to me that the existence of forward declaration necessitates that all pointers be the same size.
>
>I suppose you wouldn't lose information when casting, as long as it's guaranteed that sizeof(void*) >= sizeof(any other pointer).  Still, my understanding has always been that all pointers are the same size.
> 
>
>  
>
>>-----Original Message-----
>>From: Gareth Lewin [mailto:GL...@cl...]
>>Sent: Friday, June 27, 2003 1:45 AM
>>To: gam...@li...
>>Subject: RE: [GD-General] meaning of sizeof(int) on all plateform
>>
>>
>>btw, sizeof(pointer) isn't a real value, it is perfectly 
>>valid for a certain
>>platform to have differant sizes of pointers, so sizeof (void*) is not
>>always == sizeof(int*).
>>    
>>