virtual classes & polymorphism. example (revisited) u we want to implement a graphics system u...

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. Virtual Classes & Polymorphism

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Virtual Classes & Polymorphism

Example (revisited)

We want to implement a graphics system We plan to have lists of shape. Each shape

should be able to draw it self, compute its size, etc.

Solution #1class Shape {

public:

enum Type { Square, Circle, Triangle };

Shape( Type t, Point Center,

double width, double height);

void Draw();

double Area();

private:

Type m_type;

Point m_center;

};

Solution #1

This solution gives a nice “wrapping” to the solution we consider in C

It does not solve the problems of that solution Lack of extendibility

Solution #2 – different classesclass Square {

public:

void Draw();

double Area();

};

class Circle {

public:

void Draw();

double Area();

};

Solution #2 – Discussion

Each shape has its own implementation We can easily add new shapes

However, Shapes are different types One cannot view them as part of the same

class

Solution #3 - hierarchyclass Shape {

public:

void Draw();

double Area();

};

class Square: public Shape {

public:

void Draw();

double Area();

};

class Circle: public Shape {

public:

void Draw();

double Area();

};

Solution #3

Now we can write code such as:

Shape* MyShapes[2];

MyShapes[0] = new Circle();

MyShapes[1] = new Square();

What will happen when we callMyShapes[0]->Draw();

Lets test … see Shapes.cpp.

What is going on?

When we write:Circle circle;

circle.Draw();

The compiler calls Circle::Draw()

When we write:Shape* p = new circle();

p->Draw();

The compiler calls Shape::Draw()Why? *p has type Shape

Class Hierarchyclass Base {

public:

void foo();

void bar();

};

class Derived: public Base {

public:

void bar();

};

Derived obj;

obj.foo();

Calls Base::foo() Derived does not

implement this method

obj.bar(); Calls Derived::bar() Two implementations,

the more specific is used

Inheritance & Overloading

Derived inherits the method foo() from Base

The inherited class to use methods of the base class

Derived overloads the implementation of bar()

This mechanism allows an inherited class to specialize the implementation

Static resolution

How does the compiler determine which method to call?

Static Resolution: Based on the type of the variable.

Not the type of the object! The compiler finds the most specific

implementation of the method, and calls it

Static resolution in our example

Circle* circle = new Circle(1,1,2);

Square* square = new Square(2,2,1);

Shape* MyShapes[2];

MyShapes[0] = circle;

MyShapes[1] = square;

circle->Draw(); // Calls Circle::Draw()

square->Draw(); // Calls Square::Draw()

MyShapes[0]->Draw(); // Calls Shape::Draw()

MyShapes[1]->Draw(); // Calls Shape::Draw()

Dynamic Resolution

This clearly not what we want to in this example

More desirable here is

Dynamic resolution: Based on the type of the object Determined at run time

[Java Like]

Dynamic Resolution in C++

The virtual keyword states that the method can be overloaded in a dynamic manner.

class Base {

public:

virtual void bar();

}

class Derived: public Base {

public:

virtual void bar();

}

Solution #4: dynamic resolution

Returning to the shapes example, using virtual methods gives the desired result

See Shapes-virtual.cpp

Virtual Methods

Class Base defines a virtual method foo() The resolution of foo() is dynamic in all

subclasses of Base If the subclass Derived overloads foo(),

then Derived::foo() is called If not, Base::foo() is called

[See nested.cpp]

Underneath the Hood: Inheritance

class Base {

private:

double m_x;

int m_a;

};

class Derived

: public Base {

private:

double m_z;

};

class Base a;

class Derived b;

m_x

m_a

m_z

m_x

m_a

a:

b:Base

Pointing to an Inherited Class

class Derived b;

class Base* p = &b;

When using *p, we treat b as though it was a Base object

The compiler cannot know if *p is from a derived class or not.

m_x

m_a

m_z

b:

p:

Implementation of Virtual Methods

Possible approach: If foo() is a virtual method, then each object

has a pointer to the implementation of foo() that it uses

Essentially the solution we used in our C implementation

Cost: Each virtual method requires a pointer

Large number of virtual methods

waste of memory

Implementation of Virtual Methods

Alternative solution: Each object has a pointer to an array of

function pointer This array points to the appropriate functions

Cost: For each class, we store one table Each object contains one field that points to

the right table

Exampleclass A { public: virtual void f1(); virtual void f2(); int m_a;};

class B: public A { public: virtual void f1(); virtual void f3(); void f4();

int m_b;};… A a1, a2; B b;

class A { public: virtual void f1(); virtual void f2(); int m_a;};

class B: public A { public: virtual void f1(); virtual void f3(); void f4();

int m_b;};… A a1, a2; B b;

void A::f1{ //...};

void A::f1{ //...};

void A::f2{ //...};

void A::f2{ //...};

void B::f1{ //...};

void B::f1{ //...};

VTBLs

A

B

<vtbl>

m_a

a1:

<vtbl>

m_a

m_b

b:

<vtbl>

m_a

a1:

void B::f3{ //...};

void B::f3{ //...};

f1

f2

f3

f1

f2

Virtual - Recap

Virtual controls whether to use static or dynamic resolution

Once a method is virtual, it must remain so throughout the hierarchy

Calling a virtual method is more expensive than standard calls Two pointers are “chased” to get to the

address of the function

Destructors & Inheritance

class Base {

public:

~Base();

};

class Derived : public Base {

public:

~Derived();

};

Base *p = new Derived;

delete p;

Which destructor is called? [see destruct.cpp]

Virtual Destructor

Destructor is like any other method The example uses static resolution, and

hence the wrong destructor is called To fix that, we need to declare virtual

destructor at the base class! See destruct-virt.cpp

Once you declare virtual destructor, derived class must declare a destructor

Polymorphism & Inheritance

C++ allows to use class hierarchy to implement polymorphic code

Points of care: Choice of virtual methods

Run time considerations Use of virtual destructor for base class

Shapes & Sizes

Revisiting our example, we writeclass Shape {

public:

virtual ~Shape(); // Virtual destructor

virtual void Draw(); // Virtual method

virtual double Area(); // Also

};

How do we implement Shape::Draw() ?

Inheritance & Interfaces In this example, we never want to deal with

objects of type Shape Shape serves the role of an interace

All shapes need to be specific shapes that are instances of derived classes of Shape

How do we enforce this?Simple mechanism:void Shape::Draw(){assert(false); // we should never call this method

}

Pure Virtual Classes

We can specify that Shape::Draw() does not exist

class Shape {

public:

virtual ~Shape(); // Virtual destructor

virtual void Draw() = 0; // pure virtual

virtual double Area() = 0; // Also

};

Pure Virtual

We cannot create objects of a Pure Virtual class

Shape* p; // legal

Shape s; // illegal

Circle c; // legal

p = &c; // legal

p = new Shape; // illegal

Virtual Methods - Tips

If you have virtual methods in a class, always declare its destructor virtual

Never call virtual methods during construction and destruction

Use virtual methods Duplicate() and Create() for implementing “virtual constructors”

Use pure virtual classes to define interfaces Use inheritance with care: Be sure that this

is the appropriate solution

Post Quiz issues….

Things to brush up on: Preprocessor vs. Compiler #define and how it works Compiler vs. Linkage

Example void main()

{

(1) int *y = (int*)malloc( 3*sizeof(int) );

(2) int x[] = {1,2,3};

(3) int ** pp = NULL;

(4) int *z = y;

}

A space on the memory heap was allocated in the following lines:

a. 1

b. 2

c. 3

d. 4

Another

int *ip = (int*)malloc(100*sizeof(int));

(*ip) += 30;

free(ip);

1 This code will generate a compilation error.

2 This code can generate unpredictable run-time behavior.

3 No error would occur if we only check if ip equals NULL before calling the "free" function

4 The code contains no error.

Something else….

char s[] = {'p','l','a','b'};

What will be returned by calling strcmp(s,"plab")?

1. a negative number

2. 0

3. a positive number

4. We cannot know for sure

Almost Last Examplestruct MyStruct {

int x[3];

};

void f(MyStruct s)

{

s.x[0] = 15; s.x[1] = 16; s.x[2] = 17;

}

int main()

{

MyStruct s;

f(s);

printf("%d %d %d\n", s.x[0], s.x[1], s.x[2]);

}

Similar but not quite…struct MyStruct {

int x[3];

};

void f(MyStruct* s)

{

s->x[0] = 15; s->x[1] = 16; s->x[2] = 17;

}

int main()

{

MyStruct s;

f(&s);

printf("%d %d %d\n", s.x[0], s.x[1], s.x[2]);

}

Final Exampletypedef void (*myfunc)(int*);

void foo(int * x) {

x[0] += 2;

}

void bar( int * x ) {

x[1] += 2;

}

main()

{

myfunc funcs[2] = {

foo,

bar

};

int arr[] = {3,4,5};

(funcs[0])(arr);

(funcs[1])(arr+1);

printf("%d %d %d\n",arr[0],arr[1],arr[2]);

}