Alexander Grund wrote:
Am 03.12.24 um 12:47 schrieb Ivan Matek via Boost:
And you never do this
boost::hash2::hash_append( h, {}, v.x ); boost::hash2::hash_append( h, {}, v.y ); boost::hash2::hash_append( h, {}, v.z );
You do this instead
boost::hash2::hash_append( h, f, v.x ); boost::hash2::hash_append( h, f, v.y ); boost::hash2::hash_append( h, f, v.z );
Yes, but then again I am passing f multiple times. I don't want it to sound like I am making a big deal out of this, I can just do something like(and I actually use this idiom a lot) const auto append = [&] (const auto& val){hash_append (h, f, val);} and use that "partially applied" helper, in case I actually need to do hash_append multiple times. I think there is space for such a helper: boost::hash2::hasher append(h, f); // Or ...::hasher append(h) for default flavor append(v.x); append(v.y); append(v.z);
The usual approach here is to make hash_append variadic, so that the above becomes boost::hash2::hash_append( h, f, v.x, v.y, v.z ); I think that this is optimizing the wrong thing, and have never understood the aversion to spelling this out (it's just two Ctrl+V keystrokes), but people seem to prefer it strongly for some reason. If you define your operator== as return v1.x == v2.x && v1.y == v2.y && v1.z == v2.z; (one member per line), it's better to make tag_invoke one call per line as well, so that the two match visually and you can see the discrepancies. And if you use the `auto rep()` idiom, there's no repetition needed in either. Although, now that I think of it, if your operator== is return std::tie(v1.x, v1.y, v1.z) == std::tie(v2.x, v2.y, v2.z); then it would be better for tag_invoke to use a variadic hash_append, so that the same (v1.x, v1.y, v1.z) list is used in both. So I suppose I'll be adding the variadic overload, after all.