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#include <iostream>
#include <type_traits>
#include <vector>
 
 
class ThingBase {
public:
	virtual void printHi() = 0;
 
};
 
class Thing : public ThingBase
{
	void printHi(){
		std::cout << "hi\n";
	}	
};
 
 
template<typename ThingType>
class Container{
private:
	std::vector<ThingType> m_things;
public:
 
	typename std::enable_if<std::is_base_of<ThingBase, ThingType>::value>::type p()
	{
		m_things[0].printHi();
	};
};
 
 
int main() {
 
	//Container<Thing> stuff; // works!
	Container<int> stuff; // doesn't work :(
 
	return 0;
}

prog.cpp: In instantiation of ‘class Container<int>’:
prog.cpp:36:17:   required from here
prog.cpp:26:78: error: no type named ‘type’ in ‘struct std::enable_if<false, void>’
  typename std::enable_if<std::is_base_of<ThingBase, ThingType>::value>::type p()
                                                                              ^

Standard output is empty