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 a structured binding, then decltype yields the referenced type (described in the specification of the structured binding declaration).
|
(since C++17) |
2) If the argument is an unparenthesized id-expression naming a non-type template parameter, then decltype yields the type of the template parameter (after performing any necessary type deduction if the template parameter is declared with a placeholder type).
|
(since C++20) |
T
, andT&
;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 that prvalue. |
(until C++17) |
If expression is a prvalue other than a (possibly parenthesized) immediate invocation (since C++20), a temporary object is not materialized from that prvalue. |
(since C++17) |
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
Example
#include <iostream> struct A { double x; }; const A* a; 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 // return type can be deduced since C++14 { return t+u; } 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) |