On 16 Sep 2025 17:02, Nate Eldredge via Std-Discussion wrote:
>
>> On Sep 16, 2025, at 07:53, Nate Eldredge via Std-Discussion <std-
>> discussion_at_[hidden]> wrote:
>>
>>> On Sep 16, 2025, at 04:33, Andrey Semashev via Std-Discussion <std-
>>> discussion_at_[hidden]> wrote:
>>>
>>> On 15 Sep 2025 16:48, David Brown via Std-Discussion wrote:
>>>>
>>>> First ask yourself, is there any reason why you would want to compare
>>>> unrelated pointers for ordering?
>>>
>>> One common use case is testing whether the pointers point to a subobject
>>> of the same object. For example, if a pointer points within a string or
>>> an array. This check is often needed to properly implement operations on
>>> the string/array, such as selecting the direction of element iteration.
>>
>> I don't actually think you can do this even with std::less. It's
>> defined to use the implementation-defined strict total order over
>> pointers (https://eel.is/c++draft/defns.order.ptr <https://eel.is/c+
>> +draft/defns.order.ptr>); that has to be consistent with the <
>> operator, but no other promises are made. So for instance, given
>> `char foo[100];`, if we have `std::greater_equal<char *>{}(foo, p) &&
>> std::less<char *>{}(p, foo+3)`, I don't think the standard allows us
>> to conclude that `p == foo || p == foo+1 || p == foo+2`. The converse
>> holds, of course.
>
> That should have been `std::greater_equal<char *>{}(p, foo) &&
> std::less<char *>{}(p, foo+3)`.

Does the standard provide additional guarantees for
std::greater_equal<T*>? I thought it only did for std::less<T*>. Which
means you'd have to write `!std::less<char *>{}(p, foo) &&
std::less<char *>{}(p, foo+3)`.

Given this, if the condition is true, p *must* point to one of the first
three elements of foo. I don't see how it can be otherwise.