A function pointer that points to a function that takes an object of a template class with said function pointer as template argument. Possible?

Question

x__x

I want to do something like this:

typedef long (* fp)(BaseWindow< fp > & wnd, HWND hwnd, long wparam, long lparam);

But I get a compile error:

error C2065: 'fp' : undeclared identifier

Is it possible to implement this somehow?

Answer 1

No it isn't, because the type of the template parameter would include itself. This would yield to an endless recursion in the type.

If instead of the class template specialization, you accept a base-class of it, that's very possible

struct TemplateBase {

};

typedef long (*fpType)(TemplateBase&, HWND, long, long);

template<fpType FP>
struct BaseWindow : TemplateBase {

};


long sampleFunc(TemplateBase &b, HWND hwnd, long wparam, long lparam) {
  ...
}

int main() {
    BaseWindow<sampleFunc> bw;
    sampleFunc(bw, ...);
}

What do you want to do with this?

Answer 2

从本页上的相关链接部分： 如何键入定义以其自身类型的函数作为参数的函数指针？