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decltype specifier

Inspects the declared type of an entity or the type and value category of an expression.

Syntax

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

Explanation

1) If the argument is an unparenthesized id-expression naming an identifier introduced by a decomposition declaration, then decltype yields the referenced type (described in the specification of the decomposition declaration).
(since C++17)
2) If the argument is an unparenthesized id-expression or an unparenthesized class member access expression, then decltype yields the type of the entity named by this expression. If there is no such entity, or if the argument names a set of overloaded functions, the program is ill-formed.
3) If the argument is any other expression of type T, and
a) if the value category of expression is xvalue, then decltype yields T&&;
b) if the value category of expression is lvalue, then decltype yields T&;
c) if the value category of expression is prvalue, then decltype yields T.

If expression is a function call which returns a prvalue of class type or is a comma expression whose right operand is such a function call, a temporary object is not introduced for (until C++17)materialized from (since C++17) that prvalue. The class type need not be complete or have an available destructor. This rule doesn't apply to sub-expressions: in decltype(f(g())), g() must have a complete type, but f() need not.

Note that if the name of an object is parenthesized, it is treated as an ordinary lvalue expression, thus decltype(x) and decltype((x)) 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{0};
 
decltype(a->x) y;       // type of y is double (declared type)
decltype((a->x)) z = y; // type of z is const double& (lvalue expression)
 
template<typename T, typename 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) g = f; // the type of a lambda function is unique and unnamed
    i = f(2, 2);
    j = g(3, 3);
 
    std::cout << "i = " << i << ", "
              << "j = " << j << '\n';
}

Output:

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

See also

auto specifier specifies a type defined by an expression (C++11)
(C++11)
 obtains a reference to its argument for use in unevaluated context
(function template)

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