CODEWITHSUNDEEP
c++templatesinheritancefunction-pointersmember-pointers
volperossa
volperossa

Reputation: 1431

c++ template and inheritance with method pointer

if I declare:

class Avoidance : public Schema<std_msgs::String,prog1::Command>{

and I try to

    void*(Schema<std_msgs::String,prog1::Command>::*pt)();
    pt=&Avoidance::frontBusy;

compiler report me

error: cannot convert ‘void* (Avoidance::*)()’ 
to 
‘void* (Schema<std_msgs::String_<std::allocator<void> >, prog1::Command_<std::allocator<void> > >::*)()’ in assignment

why? Avoidance inherits from

  Schema<std_msgs::String,prog1::Command>

then Avoidance IS Schema<.....>

Upvotes: 2

Views: 191

Answers (2)

aschepler
aschepler

Reputation: 72271

Getting rid of templates to simplify, suppose you have

class B {
public:
    void f();
};

class D : public B {
public:
    void g();
};

It might seem a little backwards at first, but you can cast void (B::*)() to void (D::*)(), but you cannot cast void (D::*)() to void (B::*)(). This makes sense when you think about how they would later be used.

void test() {
    void (D::*p)() = &B::f; // OK!
    void (B::*q)() = &D::g; // ERROR!

    B b;
    D d;

    (d.*p)(); // Calls B::f on `d`.  Okay, `B::f` is an inherited member.
    (b.*q)(); // Calls D::g on `b`?? But that's not a member of `b` at all!
}

Upvotes: 2

Kerrek SB
Kerrek SB

Reputation: 476990

That's not how member function pointers work. If frontBusy is a base function, you need to type the pointer appropriately. The dispatch will still work as expected, though!

Here's a basic example:

struct A { virtual void f() = 0; };
struct B : A { virtual void f() { } };

void dispatch(void (A::*pf)(), A & a)
{            //    ^^^^^
  (a.*pf)();
}

int main()
{
  B x;
  dispatch(&A::f, x);  // calls x.B::f()
}  //      ^^^^^

So, in your case you want:

void (Schema<std_msgs::String,prog1::Command>::*p)()
                        = &Schema<std_msgs::String,prog1::Command>::frontBusy;

Upvotes: 3

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