Moreover, we can still access the breed eld of the node pointed to by ppet Suppose that in Software

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Dog::print( ) const;
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has been de ned as follows:
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void Dog::print( ) const {
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1 If you are not familiar with the -> operator, see the subsection of 10 entitled The -> Operator
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cout << "name: " << name << endl; cout << "breed: " << breed << endl; }
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The statement
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ppet->print( );
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will cause the following to be printed on the screen:
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name: Tiny breed: Great Dane
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This nice ouput happens by virtue of the fact that print( ) is a virtual member function (No pun intended) We have included test code in Display 157
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Display 157 Defeating the Slicing Problem (part 1 of 2)
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 //Program to illustrate use of a virtual function to defeat the slicing //problem #include <string> #include <iostream> We have made the member variables using std::string; public to keep the example simple In a using std::cout; real application they should be private using std::endl; and accessed via member functions class Pet { public: string name; virtual void print( ) const; }; class Dog : public Pet { public: string breed; virtual void print( ) const; }; int main( ) { Dog vdog; Pet vpet; vdogname = "Tiny"; vdogbreed = "Great Dane"; vpet = vdog; cout << "The slicing problem:\n";
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Keyword virtual is not needed here, but we put it here for clarity
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Display 157 Defeating the Slicing Problem (part 2 of 2)
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28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 //vpetbreed; is illegal since class Pet has no member named breed vpetprint( ); cout << "Note that it was print from Pet that was invoked\n"; cout << "The slicing problem defeated:\n"; Pet *ppet; ppet = new Pet; Dog *pdog; pdog = new Dog; pdog->name = "Tiny"; pdog->breed = "Great Dane"; These two print the same output: name: Tiny ppet = pdog; breed: Great Dane ppet->print( ); pdog->print( ); //The following, which accesses member variables directly //rather than via virtual functions, would produce an error: //cout << "name: " << ppet->name << " breed: " // << ppet->breed << endl; //It generates an error message saying //class Pet has no member named breed return 0; } void Dog::print( ) const { cout << "name: " << name << endl; cout << "breed: " << breed << endl; } void Pet::print( ) const { cout << "name: " << name << endl; }
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Note that no breed is mentioned
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The slicing problem: name: Tiny Note that it was print from Pet that was invoked The slicing problem defeated: name: Tiny breed: Great Dane name: Tiny breed: Great Dane
Pointers and Virtual Functions
Object-oriented programming with dynamic variables is a very different way of viewing programming This can all be bewildering at rst It will help if you keep two simple rules in mind: 1 If the domain type of the pointer pAncestor is an ancestor class for the domain type of the pointer pDescendant, then the following assignment of pointers is allowed:
pAncestor = pDescendant;
Moreover, none of the data members or member functions of the dynamic variable being pointed to by pDescendant will be lost 2 Although all the extra elds of the dynamic variable are there, you will need virtual member functions to access them
Pitfall
THE SLICING PROBLEM
Although it is legal to assign a derived class object to a base class variable, assigning a derived class object to a base class object slices off data Any data members in the derived class object that are not also in the base class will be lost in the assignment, and any member functions that are not defined in the base class are similarly unavailable to the resulting base class object For example, if Dog is a derived class of Pet, then the following is legal:
Dog vdog; Pet vpet; vpet = vdog;
However, vpet cannot be a calling object for a member function from Dog unless the function is also a member function of Pet, and all the member variables of vdog that are not inherited from the class Pet are lost This is the slicing problem Note that simply making a member function virtual does not defeat the slicing problem Note the following code from Display 157:
Dog vdog; Pet vpet; vdogname = "Tiny"; vdogbreed = "Great Dane"; vpet = vdog; vpetprint( );
Although the object in vdog is of type Dog, when vdog is assigned to the variable vpet (of type Pet) it becomes an object of type Pet So, vpetprint( ) invokes the version of print( ) defined in Pet, not the version defined in Dog This happens despite the fact that print( ) is virtual In order to defeat the slicing problem, the function must be virtual and you must use pointers and dynamic variables