Re: 'auto' and const references

Victor Bazarov <v.bazarov@comcast.invalid>
Fri, 03 May 2013 09:33:24 -0400
On 5/3/2013 9:16 AM, 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?

I think the type should be the same as of the expression. If the
expression gives you the const ref, then 'x' will be a const ref and
will bind to the value returned by the function. Say, 'thatFunction' is

     const double& thatFunction();

then the type of 'x' is going to be const double&.

Is there any difference to these:

   auto& x = thatFunction();

   const auto& x = thatFunction();

Only if 'thatFunction' does not return a const ref.

The declaration 'auto' works essentially like a template type deduction.
  The compiler needs to solve the "equation"

    decltype(x) == decltype(thatFunction());

in which it will find out the "missing" parts that 'auto' will supply
when the declaration of 'x' is complete.

So, for instance, if 'thatFunction()' is like before,

    decltype(x) == const double&

which gives you

    const auto& == const double&

, you can drop the 'const', you can drop the '&', so you get that the
'auto' supplies 'double'. IOW, the declaration of 'x' will be the same as

   const double& x

Now, take your example with 'auto& x = ', and you get

    auto & x = const double &

you can drop the '&' (since it's the most inner specifier), so you get
'auto' to equal 'const double'. That doesn't change the fact that 'x'
is going to be a 'const double&', however.

Catch my drift?

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