25 Sep
2020
25 Sep
'20
2:47 p.m.
Andrzej Krzemienski wrote:
My understanding of a "vocabulary type" is that it should be usable (not necessarily with maximum efficiency) for *any* usage.
This is not at all what a vocabulary type is. A vocabulary type is a type via which two libraries can communicate, without that type being defined by either of them. E.g. std::size_t is a vocabulary type. It's obviously not usable for *any* usage.