You know all pointers are the same size because they're all
static_cast-able to void* and back.

On Mon, Jan 27, 2025 at 9:12 AM Jeremy Rifkin via Std-Proposals <
std-proposals_at_[hidden]> wrote:

> > It's possible if the compiler hides it from the programmer. For instance
> if CHAR_BIT==8, and if the CPU instructions deal with 16-Bit operands, then
> you can have:
> >
> > CHAR_BIT == 8
> > sizeof(int*) == 2
> > sizeof(char*) == 3
>
> All the language cares about is that you can in fact address every byte of
> memory. From a language perspective, if CHAR_BIT == 8 then the smallest
> addressable unit of memory is 8 bits irrespective of hardware details.
>
> Between https://eel.is/c++draft/basic.memobj#intro.memory-1 and
> https://eel.is/c++draft/basic.compound#7 and the fact struct Frog; Frog*
> ptr1 = nullptr; Frog* foo(); Frog* ptr2 = foo(); has to work I think it's
> sufficiently covered by the standard that all pointers to non-fundamental
> objects are the same size (or at least compatible in some way to make
> everything here work). Even if you had a system where sizeof(char*) == 3
> and sizeof(int*) == 2 you would know sizeof(void*) >= 3.
>
> Cheers,
> Jeremy
>
>
> On Sun, Jan 26, 2025 at 5:42 PM Frederick Virchanza Gotham via
> Std-Proposals <std-proposals_at_[hidden]> wrote:
>
>> On Sun, Jan 26, 2025 at 11:17 PM Jeremy Rifkin wrote:
>> >
>> > > For example on a system where CHAR_BIT==8 but the smallest
>> addressable unit of memory is 16 bits
>> >
>> > That's not possible https://eel.is/c++draft/basic.memobj#intro.memory-1
>>
>>
>> It's possible if the compiler hides it from the programmer. For
>> instance if CHAR_BIT==8, and if the CPU instructions deal with 16-Bit
>> operands, then you can have:
>>
>> CHAR_BIT == 8
>> sizeof(int*) == 2
>> sizeof(char*) == 3
>>
>> An int* is a 16-Bit number representing a memory address.
>> A char* is a 16-Bit number representing a memory address, with an
>> extra byte to indicate the offset.
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>>
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