std::is_compound - cppreference.com
From cppreference.com
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(since C++11) | |
std::is_compound is a UnaryTypeTrait.
If T is a compound type (that is, array, function, object pointer, function pointer, member object pointer, member function pointer, reference, class, union, or enumeration, including any cv-qualified variants), provides the member constant value equal true. For any other type, value is false.
If the program adds specializations for std::is_compound or std::is_compound_v, the behavior is undefined.
Template parameters
Helper variable template
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(since C++17) | |
Inherited from std::integral_constant
Member constants
true if T is a compound type, false otherwise (public static member constant) |
Member functions
converts the object to bool, returns value (public member function) | |
returns value (public member function) |
Member types
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value_type
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bool
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type
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std::integral_constant<bool, value>
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Notes
Compound types are the types that are constructed from fundamental types. Any C++ type is either fundamental or compound.
Possible implementation
template<class T> struct is_compound : std::integral_constant<bool, !std::is_fundamental<T>::value> {};
Example
#include <type_traits> #include <iostream> static_assert(not std::is_compound_v<int>); static_assert(std::is_compound_v<int*>); static_assert(std::is_compound_v<int&>); void f(); static_assert(std::is_compound_v<decltype(f)>); static_assert(std::is_compound_v<decltype(&f)>); static_assert(std::is_compound_v<char[100]>); class C {}; static_assert(std::is_compound_v<C>); union U {}; static_assert(std::is_compound_v<U>); enum struct E { e }; static_assert(std::is_compound_v<E>); static_assert(std::is_compound_v<decltype(E::e)>); struct S { int i : 8; int j; void foo(); }; static_assert(not std::is_compound_v<decltype(S::i)>); static_assert(not std::is_compound_v<decltype(S::j)>); static_assert(std::is_compound_v<decltype(&S::j)>); static_assert(std::is_compound_v<decltype(&S::foo)>); int main() { std::cout << "All checks have passed\n"; }