decltype specifier (since C++11)
Inspects the declared type of an entity or the type and value category of an expression.
Syntax
decltype ( entity )
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(1) | ||||||||
decltype ( expression )
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(2) | ||||||||
Explanation
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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) |
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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). The type is non-const even if the entity is a template parameter object (which is a const object). |
(since C++20) |
T, and|
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) |
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If expression is a prvalue other than a (possibly parenthesized) immediate invocation(since C++20), a temporary object is not materialized from that prvalue: such prvalue has no result object. |
(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.
Notes
| Feature-test macro | Value | Std | Feature |
|---|---|---|---|
__cpp_decltype |
200707L | (C++11) | decltype |
Keywords
Example
#include <cassert> #include <iostream> #include <type_traits> 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; } const int& getRef(const int* p) { return *p; } static_assert(std::is_same_v<decltype(getRef), const int&(const int*)>); auto getRefFwdBad(const int* p) { return getRef(p); } static_assert(std::is_same_v<decltype(getRefFwdBad), int(const int*)>, "Just returning auto isn't perfect forwarding."); decltype(auto) getRefFwdGood(const int* p) { return getRef(p); } static_assert(std::is_same_v<decltype(getRefFwdGood), const int&(const int*)>, "Returning decltype(auto) perfectly forwards the return type."); // Alternatively: auto getRefFwdGood1(const int* p) -> decltype(getRef(p)) { return getRef(p); } static_assert(std::is_same_v<decltype(getRefFwdGood1), const int&(const int*)>, "Returning decltype(return expression) also perfectly forwards the return type."); int main() { int i = 33; decltype(i) j = i * 2; static_assert(std::is_same_v<decltype(i), decltype(j)>); assert(i == 33 && 66 == j); auto f = [i](int av, int bv) -> int { return av * bv + i; }; auto h = [i](int av, int bv) -> int { return av * bv + i; }; static_assert(!std::is_same_v<decltype(f), decltype(h)>, "The type of a lambda function is unique and unnamed"); decltype(f) g = f; std::cout << f(3, 3) << ' ' << g(3, 3) << '\n'; }
Output:
42 42
References
| Extended content |
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| This section is incomplete Reason: Requires correction. See: Talk: Wrong References. |
See also
auto specifier (C++11)
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specifies a type deduced from an expression |
| (C++11) |
obtains a reference to an object of the template type argument for use in an unevaluated context (function template) |
| (C++11) |
checks if two types are the same (class template) |
| C documentation for typeof
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