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From: Peter Dimov (pdimov_at_[hidden])
Date: 2020-09-25 14:47:04

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.

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