decltype (C++)

The decltype type specifier yields the type of a specified expression. The decltype type specifier, together with the auto keyword, is useful primarily to developers who write template libraries. Use auto and decltype to declare a template function whose return type depends on the types of its template arguments. Or, use auto and decltype to declare a template function that wraps a call to another function, and then returns the return type of the wrapped function.

Syntax

decltype( expression )

Parameters

Parameter Description
expression An expression. For more information, see Expressions.

Return Value

The type of the expression parameter.

Remarks

The decltype type specifier is supported in Visual Studio 2010 or later versions, and can be used with native or managed code. decltype(auto) (C++14) is supported in Visual Studio 2015 and later.

The compiler uses the following rules to determine the type of the expression parameter.

  • If the expression parameter is an identifier or a class member access, decltype(expression) is the type of the entity named by expression. If there is no such entity or the expression parameter names a set of overloaded functions, the compiler yields an error message.

  • If the expression parameter is a call to a function or an overloaded operator function, decltype(expression) is the return type of the function. Parentheses around an overloaded operator are ignored.

  • If the expression parameter is an rvalue, decltype(expression) is the type of expression. If the expression parameter is an lvalue, decltype(expression) is an lvalue reference to the type of expression.

The following code example demonstrates some uses of the decltype type specifier. First, assume that you have coded the following statements.

int var;
const int&& fx();
struct A { double x; }
const A* a = new A();

Next, examine the types that are returned by the four decltype statements in the following table.

Statement Type Notes
decltype(fx()); const int&& An rvalue reference to a const int.
decltype(var); int The type of variable var.
decltype(a->x); double The type of the member access.
decltype((a->x)); const double& The inner parentheses cause the statement to be evaluated as an expression instead of a member access. And because a is declared as a const pointer, the type is a reference to const double.

Decltype and Auto

In C++14, you can use decltype(auto) with no trailing return type to declare a template function whose return type depends on the types of its template arguments.

In C++11, you can use the decltype type specifier on a trailing return type, together with the auto keyword, to declare a template function whose return type depends on the types of its template arguments. For example, consider the following code example in which the return type of the template function depends on the types of the template arguments. In the code example, the UNKNOWN placeholder indicates that the return type cannot be specified.

template<typename T, typename U>
UNKNOWN func(T&& t, U&& u){ return t + u; };

The introduction of the decltype type specifier enables a developer to obtain the type of the expression that the template function returns. Use the alternative function declaration syntax that is shown later, the auto keyword, and the decltype type specifier to declare a late-specified return type. The late-specified return type is determined when the declaration is compiled, instead of when it is coded.

The following prototype illustrates the syntax of an alternative function declaration. Note that the const and volatile qualifiers, and the throw exception specification are optional. The function_body placeholder represents a compound statement that specifies what the function does. As a best coding practice, the expression placeholder in the decltype statement should match the expression specified by the return statement, if any, in the function_body.

auto function_name ( parametersopt ) constopt volatileopt -> decltype( expression ) throwopt { function_body };

In the following code example, the late-specified return type of the myFunc template function is determined by the types of the t and u template arguments. As a best coding practice, the code example also uses rvalue references and the forward function template, which support perfect forwarding. For more information, see Rvalue Reference Declarator: &&.

//C++11
template<typename T, typename U>
auto myFunc(T&& t, U&& u) -> decltype (forward<T>(t) + forward<U>(u))
        { return forward<T>(t) + forward<U>(u); };

//C++14
template<typename T, typename U>
decltype(auto) myFunc(T&& t, U&& u)
        { return forward<T>(t) + forward<U>(u); };

Decltype and Forwarding Functions (C++11)

Forwarding functions wrap calls to other functions. Consider a function template that forwards its arguments, or the results of an expression that involves those arguments, to another function. Furthermore, the forwarding function returns the result of calling the other function. In this scenario, the return type of the forwarding function should be the same as the return type of the wrapped function.

In this scenario, you cannot write an appropriate type expression without the decltype type specifier. The decltype type specifier enables generic forwarding functions because it does not lose required information about whether a function returns a reference type. For a code example of a forwarding function, see the previous myFunc template function example.

Example

The following code example declares the late-specified return type of template function Plus(). The Plus function processes its two operands with the operator+ overload. Consequently, the interpretation of the plus operator (+) and the return type of the Plus function depends on the types of the function arguments.

// decltype_1.cpp
// compile with: cl /EHsc decltype_1.cpp

#include <iostream>
#include <string>
#include <utility>
#include <iomanip>

using namespace std;

template<typename T1, typename T2>
auto Plus(T1&& t1, T2&& t2) ->
   decltype(forward<T1>(t1) + forward<T2>(t2))
{
   return forward<T1>(t1) + forward<T2>(t2);
}

class X
{
   friend X operator+(const X& x1, const X& x2)
   {
      return X(x1.m_data + x2.m_data);
   }

public:
   X(int data) : m_data(data) {}
   int Dump() const { return m_data;}
private:
   int m_data;
};

int main()
{
   // Integer
   int i = 4;
   cout <<
      "Plus(i, 9) = " <<
      Plus(i, 9) << endl;

   // Floating point
   float dx = 4.0;
   float dy = 9.5;
   cout <<
      setprecision(3) <<
      "Plus(dx, dy) = " <<
      Plus(dx, dy) << endl;

   // String
   string hello = "Hello, ";
   string world = "world!";
   cout << Plus(hello, world) << endl;

   // Custom type
   X x1(20);
   X x2(22);
   X x3 = Plus(x1, x2);
   cout <<
      "x3.Dump() = " <<
      x3.Dump() << endl;
}
Plus(i, 9) = 13
Plus(dx, dy) = 13.5
Hello, world!
x3.Dump() = 42

Example

Visual Studio 2017 and later: The compiler parses decltype arguments when the templates are declared rather than instantiated. Consequently, if a non-dependent specialization is found in the decltype argument, it will not be deferred to instantiation-time and will be processed immediately and any resulting errors will be diagnosed at that time.

The following example shows such a compiler error that is raised at the point of declaration:

#include <utility>
template <class T, class ReturnT, class... ArgsT> class IsCallable
{
public:
   struct BadType {};
   template <class U>
   static decltype(std::declval<T>()(std::declval<ArgsT>()...)) Test(int); //C2064. Should be declval<U>
   template <class U>
   static BadType Test(...);
   static constexpr bool value = std::is_convertible<decltype(Test<T>(0)), ReturnT>::value;
};

constexpr bool test1 = IsCallable<int(), int>::value;
static_assert(test1, "PASS1");
constexpr bool test2 = !IsCallable<int*, int>::value;
static_assert(test2, "PASS2");

Requirements

Visual Studio 2010 or later versions.

decltype(auto) requires Visual Studio 2015 or later.