Deep copy of a pointer function

Question

I have a class (myClass) that has a pointer to function as private member like this:

void (gi::myClass::*ptrFunction) (int  , int  , vector<int> & , vector<int> & , vector<int> &);

When you create at runtime (with new) a object (a *myClass ptrObject) of myClass this pointer is inizialited in the constructor with the address of a function. After with the same object created above I call another function. In this function I must create a deep copy of the calling object:

gi::myClass *ptrObject2 = new myClass(....);

Above I call a constructor that do the copy but it don't initialize the member *ptrFunction of *ptrObject2 My question is: How do I do a deep copy of the pointer function from the *ptrObject to ptrObject2? (My compiler is C++11)

Pointer to member function has no difference between 'deep copy' and 'shallow copy' ... there is no associated object, it's basically just an offset — M.M, Feb 17 '16 at 00:52
Namely can I do ptrObject2->ptrFunction = ptrObject->ptrFunction ? @M.M — Nick, Feb 17 '16 at 00:57
Yes I must copy a function pointer (pointer member of ptrObject to a pointer member of ptrObject2) @Ben — Nick, Feb 17 '16 at 01:05

score 3 · Answer 1 · answered Feb 17 '16 at 01:32

3

There is no deep-copy of a function pointer. You are not copying any data, just a reference into the v-table. As such:

ptrObject2->ptrFunction = ptrObject->ptrFunction

Is fine.

answered Feb 17 '16 at 01:32

Fantastic Mr Fox

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Clear. Thanks so much – Nick Feb 17 '16 at 01:33

Deep copy of a pointer function

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