AW: AW: Feature request for "friendship" of pointers in "C"

[aotto]

Hi,

there is an easy test, if Stefan is right and I'm wrong.

1. cast an INVALID Pointer to "aObj"
2. call "method_base(&aObj->obj,...)"

-> if stefan is right than "method_base" is called with an INVALID aObj 
as Pointer
-> if I am right than your software will CRASH :-)

that is the reason why I require the "frienddef".


mfg AO



On 03.02.20 09:16, aotto wrote:
> Hi,
>
> Can someone confirm that *aObj->obj does NOT create any extray "code" 
> and does NOT do any
> POINTER resolution? -> if so than "gcc" already uses the "friend" 
> logic I require :-)
>
> → I don't think that "gcc" do this already 

Hi, the following scenario has a "definition hole" in the "C" language

code example:

-------------------------
struct base {
    ...
};

struct A {
    struct base obj;
    ...
} aObj;

struct B {
    struct base obj;
    ...
} bObj;

void method_base (struct base * hdl, ...);

method_base(&aObj, ...)
method_base(&bObj, ...)
------------------------


-----Ursprüngliche Nachricht-----
Von: aotto <aotto1968@t-online.de>
Gesendet: Montag, 3. Februar 2020 09:02
An: Stefan Franke <stefan@franke.ms>
Betreff: Re: AW: Feature request for "friendship" of pointers in "C"

This is not the same!

1. usually the aObj is a Pointer:
 Â Â Â Â  -> &aObj.obj => &(aObj->obj)
2. this DOUBLE indirect access creates definitely a CRASH if
 Â Â Â Â  aObj is invalid or freed
3. I use a internal a Pointer typecheck… where I can detect if
      "my" pointer is valid or freed… but a "&(aObj->obj)" you
 Â Â Â Â  don't survive.

mfg

* Please reply at the bottom.

Ad 1: there is no need for '(' ')' if it's a pointer

 Â Â Â Â &aObj->obj


Ad 2: this is not a double indirect access, the resulting address is the 
same as aObj as long obj is the first element of your struct, only the 
type changes

 Â Â  (void *)aObj == (void*)&aObj->obj

Ad 3: since the address is the same, your check will work.

Your issues aren't related to gcc - you might consider improving your C 
skills.

Cheers

Stefan