decltype specifier

 

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Inspects the declared type of an entity or queries the return type of an expression.

Syntax

decltype ( entity )	(1)	(since C++11)
decltype ( expression )	(2)	(since C++11)

Explanation

1) If the argument is either the unparenthesised name of an object/function, or is a member access expression (object.member or pointer->member), then the decltype specifies the declared type of the entity specified by this expression.

2) If the argument is any other expression of type T, then

a) if the value category of expression is xvalue, then the decltype specifies T&&

b) if the value category of expression is lvalue, then the decltype specifies T&

c) otherwise, decltype specifies T

Note that if the name of an object is parenthesised, it becomes an lvalue expression, thus decltype(arg) and decltype((arg)) are often different types.

decltype is useful when declaring types that are difficult or impossible to declare using standard notation, like lambda-related types or types that depend on template parameters.

Keywords

decltype

Example

#include <iostream>
 
struct A {
   double x;
};
const A* a = new A();
 
decltype( a->x ) x3;       // type of x3 is double (declared type)
decltype((a->x)) x4 = x3;  // type of x4 is const double& (lvalue expression)
 
template <class T, class U>
auto add(T t, U u) -> decltype(t + u); // return type depends on template parameters
 
int main() 
{
    int i = 33;
    decltype(i) j = i*2;
 
    std::cout << "i = " << i << ", "
              << "j = " << j << '\n';
 
    auto f = [](int a, int b) -> int {
       return a*b;
    };
 
    decltype(f) f2{f}; // the type of a lambda function is unique and unnamed
    i = f(2, 2);
    j = f2(3, 3);
 
    std::cout << "i = " << i << ", "
              << "j = " << j << '\n';
}

Output:

i = 33, j = 66
i = 4, j = 9

See also

declval (C++11)	obtains the type of expression in unevaluated context (function template)