Automatic type deduction and auto&& vs auto

Question

I just watched Scott Meyers Universal References in C++11 and there was one thing I could not quite understand.

I am a bit confused as to what the difference is between a auto as a "universal reference", i.e. auto&& and a regular auto, when do they differ?

Foo f;
Foo& lvr = f;

auto lvr_a = f; // Foo&
auto rvr_a = std::move(f); // Foo&& or is it Foo?

auto&& lvr_b = f; // Foo& && => Foo&
auto&& lvr_b = std::move(f); // Foo&& && => Foo&&

score 2 · Accepted Answer · edited May 23 '17 at 12:27

2

auto will decay to the value type (ie take a copy with copy constructor), whereas auto&& will preserve the reference type.

See here: C++11: Standard ref for action of `auto` on const and reference types

auto rvr_a = std::move(f); // Foo&& or is it Foo?

It's just Foo. So this is the same as:

Foo rvr_a = std::move(f); // Foo&& or is it Foo?

But note this will still call move constructor if there is one.

edited May 23 '17 at 12:27

Community

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answered Oct 10 '12 at 12:08

Andrew Tomazos

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Would `auto&& rvr_b = std::move(f)` call the move constructor? – ronag Oct 10 '12 at 12:14
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@ronag No. The net effect is a reference binding. – Luc Danton Oct 10 '12 at 12:25
@ronag: See here: http://stackoverflow.com/questions/9657708/c11-the-range-based-for-statement-range-init-lifetime – Andrew Tomazos Oct 10 '12 at 13:35

Automatic type deduction and auto&& vs auto

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