simple structs with make_unique and emplace_back

Question

Given a struct:

struct S {
    int x;
    int y;
}

Why the Standard allows us to do this:

std::vector<S> vec;
vec.emplace_back(1, 2);

But does not allow to do this:

auto ptr = std::make_unique<S>(1, 2);

?

Works fine with me http://coliru.stacked-crooked.com/a/7c54feda67017bf1 (I had to add a constructor to the struct but that's all) — Pumkko, Jul 12 '16 at 07:56
No, both are [not allowed](http://rextester.com/WBNIK94302). If you provide a ctor (`S::S(int, int)`) then both are [fine](http://rextester.com/UHFEW17433). — songyuanyao, Jul 12 '16 at 07:57

score 9 · Answer 1 · answered Jul 12 '16 at 08:26

Actually neither work.

It was decided that emplace-style construct functions in C++ std would construct with ()s not {}s. There is no strong reason why this was chosen (that I know of).

emplace_alt snd make_unique_alt could be added to std where it constructs using {} instead. (a better name should be chosen, naturally)

So the short answer is "because std says so". The medium answer is "it is a near arbitrary choice made by std, followed elsewhere to be consistent". The long answer would involve being in the room where it happened and where it was revisited: this is not a long answer.

score 5 · Accepted Answer · answered Jul 12 '16 at 07:57

5

Please check your code.

In cpp14 your example code doesn't compile: https://ideone.com/ewyHW6

Both make_unique and emplace_back are using std::forward<Args>(args)... in background, so either both or none compiles.

answered Jul 12 '16 at 07:57

paweldac

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@vladon nobody's perfect. Mistakes happen :-) – paweldac Jul 12 '16 at 08:12

score 1 · Answer 3 · answered Jun 16 '21 at 19:31

1

how about

auto ptr = std::make_unique<S>( S{1, 2} );

without defining an extra ctor with args.

answered Jun 16 '21 at 19:31

rnd_nr_gen

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simple structs with make_unique and emplace_back

3 Answers3