C++ auto vs auto&

问题

if I have a function:

Foo& Bar()
{
   return /// do something to create a non-temp Foo here and return a reference to it
}

why is this:

auto x = Bar(); /// probably calls copy ctor - haven't checked

not the same as this?

auto &x = Bar(); /// actually get a reference here

(Actually, I'd expect the second version to get a reference to a reference, which makes little sense.)

If I explicitly specified the type of x as a value or a reference, I'll get what I expect (of course). I would expect, though, that auto would compile to the return type of Bar(), which, in this case, is a reference.

Is there an implicit cast between Foo and Foo& that comes into play here?

(Spec references accepted, though I'm getting tired of reading committee-speak.)

(Second use of time machine will be making C++ pass by reference by default. With a #pragma compatibility trigger for compiling C code. ARGH.)

回答1:

The type deduction for auto works exactly the same as for templates:

• when you deduce auto you will get a value type.
• when you deduce auto& you wil get a non-const reference type
• when you deduce const auto& you will get a const reference
• when you deduce auto&& you will get
  • a non-const reference if you assign a non-const reference
  • a const reference if you assign a const reference
  • a value when you assign a temporary

回答2:

Taken directly from Herb Sutter's blog post:

auto means “take exactly the type on the right-hand side, but strip off top-level const/volatile and &/&&.”

来源：https://stackoverflow.com/questions/20832862/c-auto-vs-auto