Re: 'auto' and const references

From:

SG <sgesemann@gmail.invalid>

Newsgroups:

comp.lang.c++

Date:

Sat, 04 May 2013 19:47:40 +0200

Message-ID:

<km3hjf$fig$1@news.albasani.net>

Am 04.05.2013 16:33, schrieb Balog Pal:

On 5/3/2013 3:16 PM, Juha Nieminen wrote:

Let's say that a function returns a const reference. If I say this:

auto x = thatFunction();

what would the type of 'x' be? Will it be a const reference, thus
eliding copying, or will a copy be made?

IIRC auto will pick up type stripped of many things. If your function
returns T& or const T & or const t, auto will still be T. But you can
write auto& or better yet auto const& that works for most cases.

Is there any difference to these:

auto& x = thatFunction();

const auto& x = thatFunction();

Due to reference collapsing these will be the same if const T& was
retuned.

I don't think reference collapsing has anything to do with it. These are
just plain template argument deduction rules.

You can use decltype() instead of auto to preserve the reference, and in
C++14 decltype(auto) will be usable in some contexts.

Looks like I have to catch up on what's going on w.r.t. standardization. :)

Cheers!
SG