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From: Bo Persson <bo@bo-persson.se>
Newsgroups: comp.lang.c++
Subject: Re: Examples of current platforms/architectures where sizeof(void*) > sizeof(int*) ?
Date: Wed, 1 Sep 2021 09:57:11 +0200
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Xref: csiph.com comp.lang.c++:80945

On 2021-09-01 at 07:41, Christian Gollwitzer wrote:
> Am 01.09.21 um 07:26 schrieb Bonita Montero:
>> Am 01.09.2021 um 06:42 schrieb Juha Nieminen:
>>> Bonita Montero <Bonita.Montero@gmail.com> wrote:
>>>> Am 29.08.2021 um 20:57 schrieb Alf P. Steinbach:
>>>>> Are there any examples of current platforms/architectures where
>>>>> sizeof(void*) > sizeof(int*) ?
>>>>
>>>> Why should there be such an architecture ?
>>>> On such an architecture you won't be able
>>>> to cast the result of malloc() to an int *
>>>
>>> Why not? If void* is larger than int*, why couldn't you cast a
>>> void* to an int*?
>>
>> Because it would drop information and f.e. you won't be able to
>> access the int-arrray behind the int *.
>>
> 
> I think you misunderstand how that works. The int pointer would not 
> count in bytes, but in ints. So if you do ptr++, then it would actually 
> increase by 1 - not by 4 or 8 as it does on the usual platforms, where 
> there is no difference between int*, char*, void* on the machine code 
> level.
> 
> Casting void* to int* would be a right shift, and if the void* points 
> inside "half an int", that would be an error - but malloc would be 
> constructed such that it only ever returns an int-aligned address. 
> Because the CPU on these platforms can work with aligned ints only.
> 
>      Christian

Right, malloc(sizeof(int)) is required to return memory properly aligned 
for objects of that size. The standard exlicitly says so.

Even on current byte addressed machines, malloc(1) gets us 8- or 
16-bytes aligned allocations, partly to reduce memory fragmentation.