How to avoid move elision when passing a temporary by value?

Question

In the following code:

Widget makeWidget()
{
    return Widget();
}

void foo(Widget widget)
{
   ...
}

foo(makeWidget());

the Widget object will be always constructed in-place (inside foo function), so no move construction takes place (at least with all compilers I've tried). What are simple examples of passing a temporary to a function by value in a way that move construction will actually take place (without explicitly moving, i.e. using std::move)?

I'm curious, why do you want to avoid move ellision? – Borgleader Apr 08 '15 at 11:54 — Borgleader, Apr 08 '15 at 11:54
I just want to understand C++ better. – rubix_addict Apr 08 '15 at 12:12 — rubix_addict, Apr 08 '15 at 12:12
Are you looking for gcc flags `-fno-elide-constructors` ? – Jarod42 Apr 08 '15 at 13:16 — Jarod42, Apr 08 '15 at 13:16

score 3 · Answer 1 · answered Apr 08 '15 at 12:03

3

The criterion for elision in this case is:

when a temporary class object that has not been bound to a reference would be copied/moved to a class object with the same cv-unqualified type

So a simple way to prevent elision would be to bind it to a reference:

foo(std::move(makeWidget());

answered Apr 08 '15 at 12:03

Mike Seymour

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I had implicit moves on my mind, so no std::move. I've edited my original question. Sorry for not being specific enough. – rubix_addict Apr 08 '15 at 12:16
@rubix_addict: In that case, I've no idea what you're asking. If you pass a temporary by value to a function with the same parameter type, then the move can be elided. There are no examples of it not being elided. – Mike Seymour Apr 08 '15 at 12:38

score 0 · Answer 2 · answered Apr 08 '15 at 13:17

0

Gcc has the flag -fno-elide-constructors which disables copy elision. With that option, both of the moves will be done in your example code.

answered Apr 08 '15 at 13:17

eerorika

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How to avoid move elision when passing a temporary by value?

2 Answers2