C++ auto& vs auto

Question

When creating local variables, is it correct to use (const) auto& or auto?

e.g.:

SomeClass object;
const auto result =          


        

Answer 1

auto uses the same mechanism of type deduction as templates, the only exception that I am aware of being that of brace-init lists, which are deduced by auto as std::initializer_list, but non-deduced in a template context.

auto x = expression;

works by first stripping all reference and cv qualifiers from the type of the right hand side expression, then matching the type. For example, if you have const int& f(){...} then auto x = f(); deduces x as int, and not const int&.

The other form,

auto& x = expression

does not strip the cv-qualifiers, so, using the example above, auto& x = f() deduces x as const int&. The other combinations just add cv qualifiers.

If you want your type to be always deduced with cv-ref qualifiers, use the infamous decltype(auto) in C++14, which uses the decltype type deduction rules.

So, in a nutshell, if you want copies, use auto, if you want references, use auto&. Use const whenever you want additional const-ness.

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EDIT There is an additional use case,

auto&& x = expression;

which uses the reference-collapsing rules, same as in the case of forwarding references in template code. If expression is a lvalue, then x is a lvalue reference with the cv-qualifiers of expression. If expression is a rvalue, then x is a rvalue reference.