Thread: invoke non-static function using ::

Hello everyone,


Confused why the following code can compile when we use :: to invoke non-static function. Especially, in the sample below, is it possible to call non-static member if T is not int? Why compiler allows the following code to compile?

Code:

#include <iostream>

using namespace std;

template  <class T>
struct FooTrait {
public:
	int foo1();
};

template<>
struct FooTrait<int> {

	static int foo1() {cout << "Hello Foo1" << endl; return 0;}
};

template <class T1, class T2 = FooTrait<T1>>
struct Foo {
public:
	int foo()
	{
		T2::foo1(); // possible to call non-static member if T is not int?

		return 0;
	}
};

 int main()
 {
	 Foo<int> f;
	 f.foo();
	 return 0;
 }

thanks in advance,
George

The compiler only creates code if you instantiate the template for a certain type.
For int everything is ok. Any other type will fail.
Kurt

Thanks Kurt,


I have got your points. Do you think it is good code or having some practical value when change the static property of some methods declared in template?

Originally Posted by ZuK

The compiler only creates code if you instantiate the template for a certain type.
For int everything is ok. Any other type will fail.
Kurt

regards,
George

Thanks CornedBee,


I have got your point that you do not think it is good code to change const property in template specialization.

Originally Posted by CornedBee

No. Specializations can add extra functionality, but in general should be used to change the implementation, not the interface.

regards,
George