is_compound Class
The latest version of this topic can be found at is_compound Class.
Tests if the specified type is not fundamental.
Syntax
template <class Ty>
struct is_compound;
Parameters
Ty
The type to query.
Remarks
An instance of the type predicate holds false if the type of Ty is a fundamental type (that is, if is_fundamental<``Ty``> holds true); otherwise, it holds true. Thus, the predicate holds true if Ty is an array type, a function type, a pointer to void or an object or a function, a reference, a class, a union, an enumeration, or a pointer to non-static class member, or a cv-qualified form of one of them.
Example
// std_tr1__type_traits__is_compound.cpp
// compile with: /EHsc
#include <type_traits>
#include <iostream>
struct trivial
{
int val;
};
int main()
{
std::cout << "is_compound<trivial> == " << std::boolalpha
<< std::is_compound<trivial>::value << std::endl;
std::cout << "is_compound<int[]> == " << std::boolalpha
<< std::is_compound<int[]>::value << std::endl;
std::cout << "is_compound<int()> == " << std::boolalpha
<< std::is_compound<int()>::value << std::endl;
std::cout << "is_compound<int&> == " << std::boolalpha
<< std::is_compound<int&>::value << std::endl;
std::cout << "is_compound<void *> == " << std::boolalpha
<< std::is_compound<void *>::value << std::endl;
std::cout << "is_compound<int> == " << std::boolalpha
<< std::is_compound<int>::value << std::endl;
return (0);
}
is_compound<trivial> == true
is_compound<int[]> == true
is_compound<int()> == true
is_compound<int&> == true
is_compound<void *> == true
is_compound<int> == false
Requirements
Header: <type_traits>
Namespace: std