How to fill an array of unique_ptr?

Question

Is it possible to use std:fill to fill an array of unique_ptrs? The intention is to have distinct pointers to distinct objects which are initialized with the same parameters.

For example:

std::unique_ptr<int> ar[3];
std::fill(ar.begin(), ar.end(), make_unique_for_each_element_somehow<int>(1));

Is `std::array` along with a `std::initializer_list` an option? — πάντα ῥεῖ, Dec 28 '16 at 15:58
`std::array` is an option but the intent is not to repeat the same filling element `N` times — Danra, Dec 28 '16 at 15:59
That's why I proposed `std::initializer_list` (or simply `{}` initialiization). — πάντα ῥεῖ, Dec 28 '16 at 15:59
@πάνταῥεῖ: I wasn't aware of it either, but it took all of 10 seconds to hypothesise that there was a `std::fill` overload or sibling function that did this job. Then 10 seconds to confirm via cppreference :) — Lightness Races in Orbit, Dec 28 '16 at 16:03

score 22 · Accepted Answer · edited May 23 '17 at 12:08

22

No, but this is what std::generate is for.

Instead of being given a single value that's copied throughout the target range, std::generate is given a "generator" function that creates each value as needed.

So, probably, something like this:

std::unique_ptr<int> ar[3];
std::generate(
   std::begin(ar),
   std::end(ar),
   []() { return std::make_unique<int>(1); }
);

I haven't tried it, and don't know whether you need to fiddle with it at all in order to avoid problems stemming from non-copyability. Hopefully move semantics are enough.

(Of course, in C++11, you will need your own make_unique function.)

By the way, your .begin() and .end() were wrong (arrays don't have member functions), so (thanks to a reminder from songyuanyao) I have corrected those.

edited May 23 '17 at 12:08

Community

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answered Dec 28 '16 at 16:01

Lightness Races in Orbit

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@songyuanyao: Yes :) – Lightness Races in Orbit Dec 28 '16 at 16:04
@Quentin: OP has tagged the question C++14. She or he or it can always fall back on a homebrew `make_unique` if needed :) – Lightness Races in Orbit Dec 28 '16 at 16:42
@LightnessRacesinOrbit to be fair I added the C++14 tag after Quentin's comment :) seeing as your solution is the most concise but only works in C++14 as written. – Danra Dec 28 '16 at 19:12
@JeJo: Indeed. Write an answer! – Lightness Races in Orbit Jun 13 '18 at 10:55
1

@JeJo: Please do not post answers as comments thanks – Lightness Races in Orbit Jun 13 '18 at 10:56
How to fix this if the size was not available in compile-time? e.g. contents from a `vector`. See https://stackoverflow.com/questions/72924076/how-to-fill-in-unique-ptr-from-vector-with-run-time-size – Sohail Si Jul 09 '22 at 19:47

score 8 · Answer 2 · answered Dec 28 '16 at 16:02

8

I don't think you can do it with std::fill() but it is trivial to do with std::generate():

std::unique_ptr<int> array[10];
std::generate(std::begin(array), std::end(array),
              []{ return std::unique_ptr<int>(new int(17)); });

answered Dec 28 '16 at 16:02

Dietmar Kühl

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score 0 · Answer 3 · answered Oct 12 '17 at 15:57

0

Another solution:

std::unique_ptr<int> ar[3];

size_t i;
for (i = 0; i < elementsof(ar); ++i) {
    ar[i].reset(new int);
}

answered Oct 12 '17 at 15:57

Yoichi Hariguchi

1

How to fill an array of unique_ptr?

3 Answers3

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