C++11 "auto" semantics

Question

When I use C++11 auto, what are the rules of type deduction with regards to whether it will resolve to a value or a reference?

E.g, sometimes it is clear:

auto i = v.begin(); // Copy, begin() returns an iterator by value 

These are less clear:

const std::shared_ptr<Foo>& get_foo(); auto p = get_foo(); // Copy or reference?  static std::shared_ptr<Foo> s_foo; auto sp = s_foo; // Copy or reference?  std::vector<std::shared_ptr<Foo>> c; for (auto foo: c) { // Copy for every loop iteration? 

Answer 1

The rule is simple : it is how you declare it.

int i = 5; auto a1 = i;    // value auto & a2 = i;  // reference 

Next example proves it :

#include <typeinfo> #include <iostream>      template< typename T > struct A {     static void foo(){ std::cout<< "value" << std::endl; } }; template< typename T > struct A< T&> {     static void foo(){ std::cout<< "reference" << std::endl; } };  float& bar() {     static float t=5.5;     return t; }  int main() {     int i = 5;     int &r = i;      auto a1 = i;     auto a2 = r;     auto a3 = bar();      A<decltype(i)>::foo();       // value     A<decltype(r)>::foo();       // reference     A<decltype(a1)>::foo();      // value     A<decltype(a2)>::foo();      // value     A<decltype(bar())>::foo();   // reference     A<decltype(a3)>::foo();      // value } 

The output:

value reference value value reference value