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

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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?

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);

edited Sep 7 at 12:36

BartoszKP

26.1k106299

asked Sep 7 at 9:43

TimW

7,25612229

As I understand it, the order of the outputs of std::transform is part of its specification, and it should not change no matter the execution policy that is chosen.
â€“Â jdehesa
Sep 7 at 9:53

add a commentÂ |Â

up vote
14
down vote

favorite

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);

edited Sep 7 at 12:36

BartoszKP

26.1k106299

asked Sep 7 at 9:43

TimW

7,25612229

As I understand it, the order of the outputs of std::transform is part of its specification, and it should not change no matter the execution policy that is chosen.
â€“Â jdehesa
Sep 7 at 9:53

add a commentÂ |Â

up vote
14
down vote

favorite

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);

edited Sep 7 at 12:36

BartoszKP

26.1k106299

asked Sep 7 at 9:43

TimW

7,25612229

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);

edited Sep 7 at 12:36

BartoszKP

26.1k106299

asked Sep 7 at 9:43

TimW

7,25612229

edited Sep 7 at 12:36

BartoszKP

26.1k106299

edited Sep 7 at 12:36

BartoszKP

26.1k106299

edited Sep 7 at 12:36

BartoszKP

26.1k106299

asked Sep 7 at 9:43

TimW

7,25612229

asked Sep 7 at 9:43

TimW

7,25612229

asked Sep 7 at 9:43

TimW

7,25612229

As I understand it, the order of the outputs of std::transform is part of its specification, and it should not change no matter the execution policy that is chosen.
â€“Â jdehesa
Sep 7 at 9:53

add a commentÂ |Â

As I understand it, the order of the outputs of std::transform is part of its specification, and it should not change no matter the execution policy that is chosen.
â€“Â jdehesa
Sep 7 at 9:53

As I understand it, the order of the outputs of std::transform is part of its specification, and it should not change no matter the execution policy that is chosen.
â€“Â jdehesa
Sep 7 at 9:53

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
11
down vote

accepted

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))).

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

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

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

edited Sep 7 at 10:01

answered Sep 7 at 9:53

Evg

2,2501130

2

You also have "Unless otherwise specified, the semantics of ExecutionPolicy algorithm overloads are identical to their overloads without" in [algorithms.parallel.overloads]
â€“Â Caleth
Sep 7 at 9:55

@Caleth, thanks, added to the answer.
â€“Â Evg
Sep 7 at 9:59

Thanks, en.cppreference.com/w/cpp/algorithm/transform doesn't mention this behavior but is seems that the standard is very clear on this
â€“Â TimW
Sep 7 at 10:03

add a commentÂ |Â

up vote
2
down vote

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.

answered Sep 7 at 9:52

Swordfish

1,566415

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
11
down vote

accepted

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))).

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

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

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

edited Sep 7 at 10:01

answered Sep 7 at 9:53

Evg

2,2501130

2

You also have "Unless otherwise specified, the semantics of ExecutionPolicy algorithm overloads are identical to their overloads without" in [algorithms.parallel.overloads]
â€“Â Caleth
Sep 7 at 9:55

@Caleth, thanks, added to the answer.
â€“Â Evg
Sep 7 at 9:59

Thanks, en.cppreference.com/w/cpp/algorithm/transform doesn't mention this behavior but is seems that the standard is very clear on this
â€“Â TimW
Sep 7 at 10:03

add a commentÂ |Â

up vote
11
down vote

accepted

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))).

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

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

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

edited Sep 7 at 10:01

answered Sep 7 at 9:53

Evg

2,2501130

2

You also have "Unless otherwise specified, the semantics of ExecutionPolicy algorithm overloads are identical to their overloads without" in [algorithms.parallel.overloads]
â€“Â Caleth
Sep 7 at 9:55

@Caleth, thanks, added to the answer.
â€“Â Evg
Sep 7 at 9:59

Thanks, en.cppreference.com/w/cpp/algorithm/transform doesn't mention this behavior but is seems that the standard is very clear on this
â€“Â TimW
Sep 7 at 10:03

add a commentÂ |Â

up vote
11
down vote

accepted

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))).

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

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

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

edited Sep 7 at 10:01

answered Sep 7 at 9:53

Evg

2,2501130

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))).

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

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

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

edited Sep 7 at 10:01

answered Sep 7 at 9:53

Evg

2,2501130

edited Sep 7 at 10:01

answered Sep 7 at 9:53

Evg

2,2501130

answered Sep 7 at 9:53

Evg

2,2501130

answered Sep 7 at 9:53

Evg

2,2501130

2

You also have "Unless otherwise specified, the semantics of ExecutionPolicy algorithm overloads are identical to their overloads without" in [algorithms.parallel.overloads]
â€“Â Caleth
Sep 7 at 9:55

@Caleth, thanks, added to the answer.
â€“Â Evg
Sep 7 at 9:59

Thanks, en.cppreference.com/w/cpp/algorithm/transform doesn't mention this behavior but is seems that the standard is very clear on this
â€“Â TimW
Sep 7 at 10:03

add a commentÂ |Â

2

You also have "Unless otherwise specified, the semantics of ExecutionPolicy algorithm overloads are identical to their overloads without" in [algorithms.parallel.overloads]
â€“Â Caleth
Sep 7 at 9:55

@Caleth, thanks, added to the answer.
â€“Â Evg
Sep 7 at 9:59

Thanks, en.cppreference.com/w/cpp/algorithm/transform doesn't mention this behavior but is seems that the standard is very clear on this
â€“Â TimW
Sep 7 at 10:03

You also have "Unless otherwise specified, the semantics of ExecutionPolicy algorithm overloads are identical to their overloads without" in [algorithms.parallel.overloads]
â€“Â Caleth
Sep 7 at 9:55

@Caleth, thanks, added to the answer.
â€“Â Evg
Sep 7 at 9:59

Thanks, en.cppreference.com/w/cpp/algorithm/transform doesn't mention this behavior but is seems that the standard is very clear on this
â€“Â TimW
Sep 7 at 10:03

add a commentÂ |Â

up vote
2
down vote

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.

answered Sep 7 at 9:52

Swordfish

1,566415

add a commentÂ |Â

up vote
2
down vote

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.

answered Sep 7 at 9:52

Swordfish

1,566415

add a commentÂ |Â

up vote
2
down vote

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.

answered Sep 7 at 9:52

Swordfish

1,566415

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.

answered Sep 7 at 9:52

Swordfish

1,566415

answered Sep 7 at 9:52

Swordfish

1,566415

answered Sep 7 at 9:52

Swordfish

1,566415

answered Sep 7 at 9:52

Swordfish

1,566415

add a commentÂ |Â

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