Scope resolution instead of . operator for template class argument

Question

below given code compiles successfuly:

template <typename T, T nontype_param>
class C;

class X {
public:

  int n;
  };

int main()
{

    C<int X::*, &X::n>* c; //Here 

    return 1;
}

How the Scope resolution operator works here rather than the . operator? Are we allowed to access a non static member like this?

reference: C++ Templates The Complete Guide , Section 8.3.3 Nontype Arguments

cdhowie · Accepted Answer · 2014-12-31T16:50:14.147

Yes, this code is valid, and the source of your confusion isn't actually related to templates.

int X::* is a pointer-to-member (specifically a pointer to int data member on type X). The pointer itself does not convey an instance of the object, so you can indeed take a pointer to a non-static member. (In fact, a pointer to a static member would just be a regular pointer, not a pointer-to-member!)

You supply the object instance at the time you use the pointer, not at the time you take the pointer.

// Declare a pointer to a data member of type int on object of type X, and
// initialize it to point to the X::n member.
int X::*ptr = &X::n;

// *ptr is not valid on member pointers. Rather, we need to supply an instance:
X instance;

// Then use the member pointer dereference operator .*
(instance.*ptr) = 5; // Same as "instance.n = 5;" in this case.

Scope resolution instead of . operator for template class argument

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