bkietz commented on a change in pull request #10397:
URL: https://github.com/apache/arrow/pull/10397#discussion_r660116713
##########
File path: cpp/src/arrow/util/vector.h
##########
@@ -84,27 +84,49 @@ std::vector<T> FilterVector(std::vector<T> values,
Predicate&& predicate) {
return values;
}
-/// \brief Like MapVector, but where the function can fail.
-template <typename Fn, typename From = internal::call_traits::argument_type<0,
Fn>,
- typename To = typename
internal::call_traits::return_type<Fn>::ValueType>
-Result<std::vector<To>> MaybeMapVector(Fn&& map, const std::vector<From>& src)
{
+template <typename Fn, typename From,
+ typename To = decltype(std::declval<Fn>()(std::declval<From>()))>
+std::vector<To> MapVector(Fn&& map, const std::vector<From>& source) {
std::vector<To> out;
- out.reserve(src.size());
- ARROW_RETURN_NOT_OK(MaybeTransform(src.begin(), src.end(),
std::back_inserter(out),
- std::forward<Fn>(map)));
- return std::move(out);
+ out.reserve(source.size());
+ std::transform(source.begin(), source.end(), std::back_inserter(out),
+ std::forward<Fn>(map));
+ return out;
}
template <typename Fn, typename From,
typename To = decltype(std::declval<Fn>()(std::declval<From>()))>
-std::vector<To> MapVector(Fn&& map, const std::vector<From>& source) {
+std::vector<To> MapVector(Fn&& map, std::vector<From>&& source) {
std::vector<To> out;
out.reserve(source.size());
- std::transform(source.begin(), source.end(), std::back_inserter(out),
+ std::transform(std::make_move_iterator(source.begin()),
+ std::make_move_iterator(source.end()),
std::back_inserter(out),
std::forward<Fn>(map));
return out;
}
+/// \brief Like MapVector, but where the function can fail.
+template <typename Fn, typename From = internal::call_traits::argument_type<0,
Fn>,
+ typename To = typename
internal::call_traits::return_type<Fn>::ValueType>
Review comment:
there's not a good reason; just uniformity with getting `From` from
call_traits.
--
This is an automated message from the Apache Git Service.
To respond to the message, please log on to GitHub and use the
URL above to go to the specific comment.
To unsubscribe, e-mail: [email protected]
For queries about this service, please contact Infrastructure at:
[email protected]