is_abstract template in C++

Last Updated : 19 Nov, 2018
The std::is_abstract template of C++ STL is used to check whether the type is a abstract class type or not. It returns a boolean value showing the same. Syntax:
template < class T > struct is_abstract;
Parameter: This template contains single parameter T (Trait class) to check whether T is a abstract class type or not. Return Value: This template returns a boolean value as shown below:
  • True: if the type is a abstract class.
  • False: if the type is a non-abstract class.
Below programs illustrate the std::is_abstract template in C++ STL: Program 1: CPP
// C++ program to illustrate
// std::is_abstract template

#include <iostream>
#include <type_traits>
using namespace std;

struct gfg {
    int m;
};

struct sam {
    virtual void foo() = 0;
};

class raj : sam {
};

int main()
{
    cout << boolalpha;
    cout << "is_abstract:" << '\n';
    cout << "gfg:" << is_abstract<gfg>::value << '\n';
    cout << "sam:" << is_abstract<sam>::value << '\n';
    cout << "raj:" << is_abstract<raj>::value << '\n';
    return 0;
}
Program 2: CPP
// C++ program to illustrate
// std::is_abstract template

#include <iostream>
#include <type_traits>
using namespace std;

struct gfg {
    virtual void foo();
};

class raj {
    virtual void foo() = 0;
};

struct sam : raj {
};

class geek {
    virtual void foo();
};

int main()
{
    cout << std::boolalpha;
    cout << "is_abstract:" << '\n';
    cout << "gfg:" << is_abstract<gfg>::value << '\n';
    cout << "raj:" << is_abstract<raj>::value << '\n';
    cout << "sam:" << is_abstract<sam>::value << '\n';
    cout << "geek:" << is_abstract<geek>::value << '\n';

    return 0;
}
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