Use of auto, decltype(), and decltype(auto)

How auto deduces types

When auto sets the type of a declared variable from its initializing expression, it proceeds as follows:

  1. First, if an auto vairable is assigned from a reference, the reference is ignored in determing the type of the auto variable. For example,
int x = 10;
int& rx = x;
auto y = rx; // The type of y is 'int'
  1. Next, if, after the step above has been performed, any const and/or volatile qualifier is also ignored. For example:
int x = 10;
const int& rx = x;
*crx = 11; // error: *crx is read only
auto y = rx; // The type of y is 'int'

The type of y above is int. Both the reference and const are ignored. To make y a reference you simply use auto& or const auto&&:

int x = 10;
const int& rx = x;
auto& y = rx; // The type of y is 'int&'
const auto& a = rx; // The type of a is 'const int&'

Use of auto with const pointers

If a pointer of type some_type *const, is assigned to a variable declared auto, the type will be some_type *:

auto x = 10;       // x is 'int'
int *const p = &x; // *p is read/write, but p is read-only
*p  = 20;          // x is now 20
auto y = 50;       // y is 'int'
p = y;             // error p is const, read-only

auto pauto1 = p;       // pauto1 is 'int *'
const auto pauto2 = p; // pauto2 is 'int *const' just like p

If a pointer of type const some_type *, is assigned to a variable declared auto, the type will be const some_type *:

auto x = 10;       // x is 'int'
const int *p = &x; // *p is read/write, but p is read-only

auto pauto = p;        // pauto1 is 'const int *'
const auto pauto = p; // pauto1 is 'const int * const'

The auto deduced types for pointer involving const follow common sense rules: they preserve const when it is necessary; otherwise, they ignore it.

Further Examples

To further clarify the use of auto&. What happens when a auto& rc is assigned from a const int?

const int c = 0;
auto& rc = c;
rc = 44; // Is rc an 'int &' or an 'const int&'?

Obviously, c cannot be changed because it is const. Therefore rc must be a const int&. Note: This example still adheres to the two rules above (since c was not a reference).

Use of auto&&

auto&& behaves like template functions parameters that are declared using &&:


Provide a cross link to forwarding references here.

template<typename T> T f(T&& t)
  T ret{t};
  return ret;

class Example {
        std::vector<int> v;
       Example() : v{0, 1, 2, 3} {}
       Example(const Example& lhs) : v{lhs}
          cout << "Example copy ctor called." << std::endl;

       Example(Example&& lhs) : v{std::move(lhs}
          cout << "Example move ctor called." << std::endl;
       const std::vector<int>&  get_vector() const { return v;}

Example example1{}; // lvalue

T t1{  f(example1) };
T t2{ f(Example{} );  // rvalue

decltype(name) and decltype(expression) deduction rules

decltype means the ‘declared type’. If you use decltype with a name, it will give you the declared type of that name:

int x = 10;
decltype(x); //  decltype(x) = int

const auto& rx = x;
decltype(rx); //  decltype(x) = const int&

If you have an expression instead of a name, then decltype(expr) is either an lvalue or an rvalue. If it an lvalue, then decltype will add a reference to it. Below when we add parenthesis to x before passing it to decltype, we turn it into an expression; it is not longer solely a name:


and the result of decltype((x)) is int & because (x) is an expression not a name, and thus decltype adds a reference to the type of the lvalue expression.

Template Functions Returning auto versus decltype(auto)

Consider this function template that whose return type is declared auto

template<class Container, class Index> auto get_value(Container& c, Index i)
    return c[i];

vector<int> v{1, 2, 3 ,4, 5};

cout << "get_value(1, 3) = is: " << get_value(v, 3) << endl;

This produces the expected output of:

get_value(v, 3) is: 4

However, assigning to get_values(v, 3) = 10 fails to compile. Why? Most containers with an index operator like std::vector<int> return an reference to an lvalue: in the case of vecotr<int>, an int& is retunred. However, if the return type of the template get_value() is auto, instead of returning in&, int is returned. That is, the value of the return type is the same as the value of x below:

vector<int> v{1 ,2 ,3 ,4 5};
auto y = v[3]; // y is of type 'int' not 'int&'

This is because the auto return type uses template (not auto) type deduction rules and not the normal auto type deduction rules for objects. But, again, when auto is used as a return type, it uses template type deduction rules. Therefore, to return the desired int& return type above, the type identical to c[i], we must use decltype(auto), which will retun the same type as y and z below

vector<int> v{1, 2, 3, 4, 5};

auto x = v[3];

decltype(auto) y = v[3];

decltype(v[3]) z = v[3];

y = 10;

cout << "v[3] = " << v[3] << ", x = " << x << ", y = " << y << ", and z = " << z << endl;

The output is:

v[3] = 10, x = 4, y = 10, and z = 10

because the decltype(auto) means ‘automatically deduce the return type using the decltype type deduction rules’. So we must reimplement get_values() as

template<class Container, class Index> decltype(auto) get_value(Container& c, Index i)
    return c[i];

vector<int> v{1, 2, 3, 4, 5};

get_value(v, 3) = 10;

cout << "v[3] = " << v[3] << ", get_value(v, 3) = " << get_value(v, 3) << endl;

which produces:

v[3] = 10, get_value(v, 3) = 10


The required C++11 syntax for get_value() would have been:

template<class Container, class Index> auto get_value(Container& c, Index i) ->  decltype(c[i])
    return c[i];

In summary, we need to know the use case for your function: do you want template type deduction rules, then use auto for the return type; if you want the decltype type deduction, then use decltype(auto). It often boils down to whether you want an lvalue reference return or an rvalue. In general, decltype(auto) will return the type of the actual expression or object being returned. So in general it is the first choice to always consider. rules described above.

Finally, the same comments about template returns types apply to lambdas.

Using decltype(declval<some_type>())

The entry for decval explains:

Returns an rvalue reference to type T without referring to any object.

This function shall only be used in unevaluated operands (such as the operands of sizeof and decltype).

T may be an incomplete type.

This is a helper function used to refer to members of a class in unevaluated operands, especially when either the constructor signature is unknown or when no objects of that type can be constructed (such as for abstract base classes).

And it gives this example:

// declval example
#include <utility>      // std::declval
#include <iostream>     // std::cout

struct A {              // abstract class
  virtual int value() = 0;

class B : public A {    // class with specific constructor
  int val_;
  B(int i,int j):val_(i*j){}
  int value() {return val_;}

int main() {
  decltype(std::declval<A>().value()) a;  // int a
  decltype(std::declval<B>().value()) b;  // int b
  decltype(B(0,0).value()) c;   // same as above (known constructor)
  a = b = B(10,2).value();
  std::cout << a << '\n';
  return 0;