> On 2 Sep 2025, at 14:49, David Brown via Std-Proposals <std-proposals_at_[hidden]> wrote:
>
> On 02/09/2025 14:24, Hans Åberg via Std-Proposals wrote:
>>> On 2 Sep 2025, at 14:14, Jan Schultke <janschultke_at_[hidden]> wrote:
>>>
>>> You seem to be confusing some mostly unrelated concepts.
>>>
>>>>> 1. C does not allow _BitInt(1); should C++ to make generic programming
>>>>> more comfortable?
>>>>
>>>> The ring ℤ/2ℤ of integers modulo 2, also a field, is isomorphic to the Boolean ring 𝔹 having exclusive or as addition and logical conjunction as multiplication.
>>>>
>>>> If bool 1+1 is defined to 0, then it is already in C++.
>>>
>>> Whether there is some other C++ thing that works mathematically the
>>> same doesn't say anything about whether _BitInt(1) is valid or should
>>> be valid. The issue is regarding a specific type.
>> It could have been defined to be the same as bool.
>
> No, it could not. _BitInt(1), if it is to exist, has the two values -1 and 0. Like all signed integer types, arithmetic overflow on it is undefined, and like all _BitInt types, there is no integer promotion. Thus for _BitInt(1), (-1) + (-1) is UB.
>
> unsigned _BitInt(1) can hold 0 and 1, has no integer promotion, and has modulo arithmetic behaviour - thus 1 + 1 gives 0 for "unsigned _BitInt(1)".

Right. I was thinking about the unsigned type only.

> Both signed _BitInt(1) and unsigned _BitInt(1) are integer types representing numbers. In C23 (and presumably C++ when it gets _BitInt), these are included in the "standard integer types". "bool", on the other hand, is /not/ a standard integer type in C++ (though it is one in C).

I get bool true + true = 2 also in C, so it seems it does not have proper arithmetic + in the sense that the results are the same as in conversion back and forth to int.

> Being able to represent two distinct values does not mean a type behaves like "bool".
>
>>>> However, in GCC and Clang, bool 1+1 = 1, and I could not see what the standard specifies.
>>>
>>> Any nonzero integer is true when converted to bool, so "bool(true +
>>> true)" is essentially "1 + 1 != 0". bool gets promoted to int before
>>> any operation.
>> Apparently, there is no operator+ for bool. I did:
>> bool a = 1, b = 1, c = a + b;
>
> Correct. There is also no "+" operator for char, or short int - other than _BitInt, types smaller than "int" get promoted to "int" before the operation.

Right.

The implicit down-conversions, that lose information, get in the way here.