Meaning of "const -> std::string const&" after the function definition?

Question

Reading the answer for one exercise in C++ Primer, 5th Edition, I found this code:

#ifndef CP5_ex7_04_h
#define CP5_ex7_04_h
#include <string>

class Person {
std::string name;
std::string address;
public:

auto get_name() const -> std::string const& { return name; }
auto get_addr() const -> std::string const& { return address; }
};

#endif

What does

const -> std::string const& 

mean in this case?

Answer 1

auto get_name() const -> std::string const& { return name; } is trailing return type notation for the equivalent

std::string const& get_name() const { return name; }

Note that the equivalence is exact in the sense that you can declare a function using one syntax and define it with the other.

(This has been part of the C++ standard since and including C++11).

Answer 2

The part -> std::string const& is trailing return type and is new syntax since C++11.

The first const says it a const member function. It can be safely called on a const object of type Person.

The second part simply tells what the return type is - std:string const&.

It is useful when the return type needs to be deduced from a template argument. For known return types, it is no more useful than than using:

std::string const& get_name() const { return name; }