在本文中,我们将讨论C ++ STL中std::is_polymorphic模板的工作,语法和示例。
is_polymorphic是C ++中<type_traits>头文件下的模板。该模板用于检查该类是否为多态类,并相应地返回结果true或false。
从声明虚拟函数的抽象类中声明虚拟函数的类。该类具有在其他类中声明的虚函数的声明。
template <class T> is_polymorphic;
模板只能具有类型T的参数,并检查给定类型是否为多态类。
它返回一个布尔值,如果给定类型是多态类,则返回true,如果给定类型不是多态类,则返回false。
Input: class B { virtual void fn(){} }; class C : B {}; is_polymorphic<B>::value; Output: True Input: class A {}; is_polymorphic<A>::value; Output: False
#include <iostream> #include <type_traits> using namespace std; struct TP { virtual void display(); }; struct TP_2 : TP { }; class TP_3 { virtual void display() = 0; }; struct TP_4 : TP_3 { }; int main() { cout << boolalpha; cout << "Checking for is_polymorphic: "; cout << "\n structure TP with one virtual function : "<<is_polymorphic<TP>::value; cout << "\n structure TP_2 inherited from TP: "<<is_polymorphic<TP_2>::value; cout << "\n class TP_3 with one virtual function: "<<is_polymorphic<TP_3>::value; cout << "\n class TP_4 inherited from TP_3: "<< is_polymorphic<TP_4>::value; return 0; }
输出结果
如果我们运行上面的代码,它将生成以下输出-
Checking for is_polymorphic: structure TP with one virtual function : true structure TP_2 inherited from TP: true class TP_3 with one virtual function: true class TP_4 inherited from TP_3: true
#include <iostream> #include <type_traits> using namespace std; struct TP { int var; }; struct TP_2 { virtual void display(); }; class TP_3: TP_2 { }; int main() { cout << boolalpha; cout << "Checking for is_polymorphic: "; cout << "\n structure TP with one variable : "<<is_polymorphic<TP>::value; cout << "\n structure TP_2 with one virtual function : "<<is_polymorphic<TP_2>::value; cout << "\n class TP_3 inherited from structure TP_2: "<<is_polymorphic<TP_3>::value; return 0; }
输出结果
如果我们运行上面的代码,它将生成以下输出-
Checking for is_polymorphic: structure TP with one variable : false structure TP_2 with one virtual function : true class TP_3 inherited from structure TP_2 : true