Can I still rely on the order of the output elements when using par_unseq?

After reading the documentation I'm still confused about the usage of par_unseq. I know I can't tell anything about the order of execution because of threading and vectorization but can I still rely on the order of the outputs?

par_unseq

transform([x0, x1, x2], f) == [f(x0), f(x1), f(x2)]]


In order words, is this test ever going to fail?

std::vector<int> xs = 1, 2, 3, 4;
std::vector<int> ys(xs.size());

std::transform(
std::execution::par_unseq,
cbegin(xs), cend(xs),
begin(ys),
(int x) return x*x; );

std::vector<int> expected = 1, 4, 9, 16;
ASSERT_EQ(expected , ys);


std::transform

2 Answers
2

The Standard, [alg.transform], reads:

Effects: Assigns through every iterator i in the range [result,result + (last1 - first1)) a new corresponding value equal to op(*(first1 + (i - result)) or binary_op(*(first1 + (i - result)), *(first2 + (i - result))).

i

[result,result + (last1 - first1))

op(*(first1 + (i - result))

binary_op(*(first1 + (i - result)), *(first2 + (i - result)))

and (thanks, @Caleth), [algorithms.parallel.overloads]:

Unless otherwise specified, the semantics of ExecutionPolicy algorithm overloads are identical to their overloads without.

ExecutionPolicy

So, yes, you can rely on the order in the output.

No, your test is never going to fail because even if order of execution changes,
ys[0...3] = xs[0...3] * xs[0...3] = 1*1, 2*2, 3*3, 4*4;
won't change.

ys[0...3] = xs[0...3] * xs[0...3] = 1*1, 2*2, 3*3, 4*4;

Thanks for contributing an answer to Stack Overflow!

But avoid …

To learn more, see our tips on writing great answers.

Required, but never shown

Required, but never shown

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.