So you (kind of) say __BitInt(m) has modulus 2^m int_mod<m> has modulus m? That would be an actual difference in capability on the C++ side. Or you are just saying that even with int_mod<m> the modulus is 2^m, but the bit width of the underlying type may be different? -----Ursprüngliche Nachricht----- Von:Hans Åberg <haberg_1_at_[hidden]> Gesendet:Mi 14.01.2026 15:35 Betreff:Re: [std-proposals] Modular integers An:std-proposals_at_[hidden]; CC:Sebastian Wittmeier <wittmeier_at_[hidden]>; > On 14 Jan 2026, at 00:10, Sebastian Wittmeier via Std-Proposals <std-proposals_at_[hidden]> wrote: > > Again (in other words) a compiler can internally map between _BitInt(m) and int_mod<m> notations. The modulus is not the number of bits.