Thread: Function pointer

Isn't this a little difficult to understand ... You would think a return type is just 'float' ... And can't see two functions in this signature ..

Code:

   // function takes a char and returns a pointer to a function which is taking two
   // floats and returns a float. <opCode> specifies which function to return
   float (*GetPtr1(const char opCode))(float, float)
   {
      if(opCode == '+')  return &Plus;
      if(opCode == '-')  return &Minus;
   }

Originally Posted by matsp

No, because of the (* at the beginning of the function name and the )(float, float) after the function parameter, you have said "This is a pointer to a function returning float".

The syntax for function pointers is definitely a bit messy - but fortunately, we can do things to simplify it, by using typedef. I personally think this makes your code more readable:

Code:

typedef float (*OperatorFunc)(float, float);

   // function takes a char and returns a pointer to a function which is taking two
   // floats and returns a float. <opCode> specifies which function to return
   OperatorFunc GetPtr1(const char opCode)
   {
      if(opCode == '+')  return &Plus;
      if(opCode == '-')  return &Minus;
   }

--
Mats

Hi Mats, welcome back ...

Well, i must say a lot is hidden without typedef... you have added 'GetPtr1' which was not part of the definition i've provided ... That makes it more readable indeed!!

Say i have a member function pointer

Code:

class MyClass {
private:
       typedef void (MyClass::*MyFunct)(const string&);
       MyFunct f;

public :
      void GeneralFunct(const string& s);
}

.....

{

MyFunct f = &MyClass::GeneralFunct;

this->*f("Hello There");

/* OR */

MyClass *class1;
class1->*f("Hello There");

/* OR */

MyClass class2;
class2.*f("Hello There");

}

Assuming what i've done is correct, why do i always need a Class Pointer to invoke a member function pointer?

Why can't i just ...

Code:

       MyClass myclass = &MyClass::GeneralFunct;
       myclass.f("What's wrong with this?");

Originally Posted by Elysia

But you can call a function through a class reference...

Code:

class foo
{
public:
	void foo2() {}
};

int main()
{
	typedef void (foo::* foo_member_ptr)();
	foo_member_ptr my_ptr = &foo::foo2;
	foo f;
	(f.*my_ptr)();
	foo& f2 = f;
	(f2.*my_ptr)();
	foo* pf = &f;
	(pf->*my_ptr)();
}

It thought a reference was the same as a pointer ... I don't see any difference in what you done as to my examples ...

Originally Posted by matsp

There is no such thing as a "function reference", as functions are not variables, and a reference refers to a variable. You could have a reference to a function pointer, but you still have to make the function "variable" a pointer.

--
Mats

So my question is left unanswered !!! ... anyways, its not an issue ...

Originally Posted by matsp

There is no such thing as a "function reference", as functions are not variables, and a reference refers to a variable. You could have a reference to a function pointer, but you still have to make the function "variable" a pointer.

--
Mats

Pardon me, i got my mind mixed up somewhere ... I'm referring to a class pointer while you referring to a pointer variable to a function ...

I understand the whole referencing now .. thanx Elysia & Mats ...

@Mats, you should see the progress of that Rental Project! but now you slamming the clrscrn .. anyway ...

How useful are delegates over members function pointers?