is_pod template in C++
Last Updated :
19 Nov, 2018
The std::is_pod template of C++ STL is used to check whether the type is a plain-old data(POD) type or not. It returns a boolean value showing the same.
Syntax:
template < class T > struct is_pod;
Parameter: This template contains single parameter T (Trait class) to check whether T is a pod type or not.
Return Value: This template returns a boolean value as shown below:
- True: if the type is a pod type.
- False: if the type is a non-pod type.
Below programs illustrate the std::is_pod template in C++ STL:
Program 1:
#include <iostream>
#include <type_traits>
using namespace std;
struct gfg {
int var1;
};
struct sam {
int var2;
private :
int var3;
};
struct raj {
virtual void func();
};
int main()
{
cout << boolalpha;
cout << "is_pod:" << '\n' ;
cout << "gfg:" << is_pod<gfg>::value << '\n' ;
cout << "sam:" << is_pod<sam>::value << '\n' ;
cout << "raj:" << is_pod<raj>::value << '\n' ;
return 0;
}
|
Output:
is_pod:
gfg:true
sam:false
raj:false
Program 2:
#include <iostream>
#include <type_traits>
using namespace std;
class gfg {
int var1;
};
class sam {
int var2;
private :
int var3;
};
class raj {
virtual void func();
};
int main()
{
cout << boolalpha;
cout << "is_pod:" << '\n' ;
cout << "gfg:" << is_pod<gfg>::value << '\n' ;
cout << "sam:" << is_pod<sam>::value << '\n' ;
cout << "raj:" << is_pod<raj>::value << '\n' ;
return 0;
}
|
Output:
is_pod:
gfg:true
sam:true
raj:false
Program 3:
#include <iostream>
#include <type_traits>
using namespace std;
union gfg {
int var1;
};
union sam {
int var2;
private :
int var3;
};
int main()
{
cout << boolalpha;
cout << "is_pod:" << '\n' ;
cout << "gfg:" << is_pod<gfg>::value << '\n' ;
cout << "sam:" << is_pod<sam>::value << '\n' ;
return 0;
}
|
Output:
is_pod:
gfg:true
sam:false
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