wash added a comment.
Suppose you have:
struct A {};
struct B {};
A operator+(A, B);
std::vector<B> v;
std::reduce(v.begin(), v.end(), A{});
The implementation in this patch works fine for the above case - as would
`accumulate`, which it is equivalent to. However, `reduce` requires that "All
of `binary_op(init, *first)`, `binary_op(*first, init)`, `binary_op(init,
init)`, and `binary_op(*first, *first)` shall be convertible to T.", as all
those operations may be needed for the parallel execution-policy overload of
`reduce` (the requirement applies to the non-execution-policy overload as well).
E.g. in the above example, these operations are also required by `reduce`, even
though the non-parallel implementation does not need them:
A operator+(B, A);
A operator+(A, A);
A operator+(B, B);
Should the non-parallel implementation of `reduce` static_assert or SFINAE away
when these requirements are not met? I think it might be desirable to ensure
that if this won't compile:
std::reduce(std::par, v.begin(), v.end(), A{});
Then this will also not compile:
std::reduce(v.begin(), v.end(), A{});
(And the spec seems to suggest this too).
================
Comment at: include/numeric:154
+ return reduce(__first, __last,
+ typename iterator_traits<_InputIterator>::value_type{},
_VSTD::plus<>());
+}
----------------
In the spec, this overload of `reduce` is described as equivalent to `return
reduce(std::forward<ExecutionPolicy>(exec), first, last, typename
iterator_traits<ForwardIterator>::value_type{});`. The overload that it calls
(the three argument version that omits a binary operation) just forwards to the
four-argument reduce, adding the `plus<>()` argument.
Is there a reason you wanted to avoid the extra layer of function call
indirection (it should be inlined and optimized away, right)? What you have
seems perfectly fine, I'm just curious though.
================
Comment at: test/std/numerics/numeric.ops/reduce/reduce_iter_iter.pass.cpp:37
+ unsigned sa = sizeof(ia) / sizeof(ia[0]);
+ test(Iter(ia), Iter(ia), 0);
+ test(Iter(ia), Iter(ia+1), 1);
----------------
Just to confirm, this should be 0 because the "default" init value is
`iterator_traits<_InputIterator>::value_type{}`, and `int{}` gives you a
determinate result (as opposed to `int()` which would not), correct?
https://reviews.llvm.org/D33997
_______________________________________________
cfe-commits mailing list
[email protected]
http://lists.llvm.org/cgi-bin/mailman/listinfo/cfe-commits
