I'm reading Scott Meyers' More Effective C++ now. Edifying! Item 2 mentions that dynamic_cast can be used not only for downcasts but also for sibling casts. Could please anyone provide a (reasonably) non-contrived example of its usage for siblings? This silly test prints 0 as it should, but I can't imagine any application for such conversions.
#include <iostream>
using namespace std;
class B {
public:
virtual ~B() {}
};
class D1 : public B {};
class D2 : public B {};
int main() {
B* pb = new D1;
D2* pd2 = dynamic_cast<D2*>(pb);
cout << pd2 << endl;
}
The scenario you suggested doesn't match sidecast exactly, which is usually used for the casting between pointers/references of two classes, and the pointers/references are referring to an object of class which both derives from the two classes. Here's an example for it:
struct Readable {
virtual void read() = 0;
};
struct Writable {
virtual void write() = 0;
};
struct MyClass : Readable, Writable {
void read() { std::cout << "read"; }
void write() { std::cout << "write"; }
};
int main()
{
MyClass m;
Readable* pr = &m;
// sidecast to Writable* through Readable*, which points to an object of MyClass in fact
Writable* pw = dynamic_cast<Writable*>(pr);
if (pw) {
pw->write(); // safe to call
}
}
LIVE
It is called cross-cast, and it is used when a class inherits from two different classes (not the other way around, as shown in your question).
For example, given the following class-hierarchy:
A B
\ /
C
If you have an A pointer to a C object, then you can get a B pointer to that C object:
A* ap = new C;
B* bp = dynamic_cast<B*>(ap);
Related
I have the following structure of classes:
struct A {
virtual void function() = 0;
};
struct B : A {
void function() override {
// ...implementation...
}
};
struct C {
virtual A a() = 0;
};
struct D : public C {
A a() {
return B{};
}
}
the main idea is: I do not really care what class will be returned from D::a, the only property I want it to have is a function() member-function. So I denoted the return type as the abstract class A which has this property as virtual, and then try to return B object which is derived from A. But this does not work, giving error: invalid abstract return type. Could you please suggest a way to achieve the desired effect in C++?
In C++, you can't do polymorphism with objects, because objects could have different sizes, and in C++ types have value semantics. You can, however, do polymorphism with references, and pointers. So your function could be:
A* a();
// or
A& a();
This does mean that you may have problems with ownership. You can also return a smart pointer instead:
unique_ptr<A> a();
You can't create objects of abstract types, you can create pointers of them. As Ayxan said, you can use also smart pointers, but I suggest you use shared_ptr because it's more comfortable.
Use dynamic_cast to convert B to A.
You forgot to add ; to the end of the third struct.
Here is the correct code:
#include <iostream>
struct A {
virtual void function() = 0;
};
struct B : A {
void function() override {
std::cout << "EEAAAHHHH\n";
}
};
struct C {
virtual A* a() = 0;
};
struct D : public C {
A* a() {
B *a = new B;
A *b = dynamic_cast<A*>(a);
return b;
}
};
int main()
{
D d;
A* a = d.a();
a->function();
return 0;
}
If you compile and run the code, you can see that the output is "EEAAAHHHH".
I'm happy to help you, and I wish you success.
If I declare a base class (or interface class) and specify a default value for one or more of its parameters, do the derived classes have to specify the same defaults and if not, which defaults will manifest in the derived classes?
Addendum: I'm also interested in how this may be handled across different compilers and any input on "recommended" practice in this scenario.
Virtuals may have defaults. The defaults in the base class are not inherited by derived classes.
Which default is used -- ie, the base class' or a derived class' -- is determined by the static type used to make the call to the function. If you call through a base class object, pointer or reference, the default denoted in the base class is used. Conversely, if you call through a derived class object, pointer or reference the defaults denoted in the derived class are used. There is an example below the Standard quotation that demonstrates this.
Some compilers may do something different, but this is what the C++03 and C++11 Standards say:
8.3.6.10:
A virtual function call (10.3) uses
the default arguments in the
declaration of the virtual function
determined
by the static type of the pointer or reference denoting the object. An
overriding function in a derived
class does not acquire default arguments from the function it
overrides. Example:
struct A {
virtual void f(int a = 7);
};
struct B : public A {
void f(int a);
};
void m()
{
B* pb = new B;
A* pa = pb;
pa->f(); //OK, calls pa->B::f(7)
pb->f(); //error: wrong number of arguments for B::f()
}
Here is a sample program to demonstrate what defaults are picked up. I'm using structs here rather than classes simply for brevity -- class and struct are exactly the same in almost every way except default visibility.
#include <string>
#include <sstream>
#include <iostream>
#include <iomanip>
using std::stringstream;
using std::string;
using std::cout;
using std::endl;
struct Base { virtual string Speak(int n = 42); };
struct Der : public Base { string Speak(int n = 84); };
string Base::Speak(int n)
{
stringstream ss;
ss << "Base " << n;
return ss.str();
}
string Der::Speak(int n)
{
stringstream ss;
ss << "Der " << n;
return ss.str();
}
int main()
{
Base b1;
Der d1;
Base *pb1 = &b1, *pb2 = &d1;
Der *pd1 = &d1;
cout << pb1->Speak() << "\n" // Base 42
<< pb2->Speak() << "\n" // Der 42
<< pd1->Speak() << "\n" // Der 84
<< endl;
}
The output of this program (on MSVC10 and GCC 4.4) is:
Base 42
Der 42
Der 84
This was the topic of one of Herb Sutter's early Guru of the Week posts.
The first thing he says on the subject is DON'T DO THAT.
In more detail, yes, you can specify different default parameters. They won't work the same way as the virtual functions. A virtual function is called on the dynamic type of the object, while the default parameter values are based on the static type.
Given
class A {
virtual void foo(int i = 1) { cout << "A::foo" << i << endl; }
};
class B: public A {
virtual void foo(int i = 2) { cout << "B::foo" << i << endl; }
};
void test() {
A a;
B b;
A* ap = &b;
a.foo();
b.foo();
ap->foo();
}
you should get
A::foo1
B::foo2
B::foo1
This is a bad idea, because the default arguments you get will depend on the static type of the object, whereas the virtual function dispatched to will depend on the dynamic type.
That is to say, when you call a function with default arguments, the default arguments are substituted at compile time, regardless of whether the function is virtual or not.
#cppcoder offered the following example in his [closed] question:
struct A {
virtual void display(int i = 5) { std::cout << "Base::" << i << "\n"; }
};
struct B : public A {
virtual void display(int i = 9) override { std::cout << "Derived::" << i << "\n"; }
};
int main()
{
A * a = new B();
a->display();
A* aa = new A();
aa->display();
B* bb = new B();
bb->display();
}
Which produces the following output:
Derived::5
Base::5
Derived::9
With the aid of the explanation above, it is easy to see why. At compile time, the compiler substitutes the default arguments from the member functions of the static types of the pointers, making the main function equivalent to the following:
A * a = new B();
a->display(5);
A* aa = new A();
aa->display(5);
B* bb = new B();
bb->display(9);
As other answers have detailed, its bad idea. However since no one mentions simple and effective solution, here it is: Convert your parameters to struct and then you can have default values to struct members!
So instead of,
//bad idea
virtual method1(int x = 0, int y = 0, int z = 0)
do this,
//good idea
struct Param1 {
int x = 0, y = 0, z = 0;
};
virtual method1(const Param1& p)
As you can see from the other answers this is a complicated subject. Instead of trying to do this or understand what it does (if you have to ask now, the maintainer will have to ask or look it up a year from now).
Instead, create a public non-virtual function in the base class with default parameters. Then it calls a private or protected virtual function that has no default parameters and is overridden in child classes as needed. Then you don't have to worry about the particulars of how it would work and the code is very obvious.
This is one that you can probably figure out reasonably well by testing (i.e., it's a sufficiently mainstream part of the language that most compilers almost certainly get it right and unless you see differences between compilers, their output can be considered pretty well authoritative).
#include <iostream>
struct base {
virtual void x(int a=0) { std::cout << a; }
virtual ~base() {}
};
struct derived1 : base {
void x(int a) { std:: cout << a; }
};
struct derived2 : base {
void x(int a = 1) { std::cout << a; }
};
int main() {
base *b[3];
b[0] = new base;
b[1] = new derived1;
b[2] = new derived2;
for (int i=0; i<3; i++) {
b[i]->x();
delete b[i];
}
derived1 d;
// d.x(); // won't compile.
derived2 d2;
d2.x();
return 0;
}
I have a base class instance, there is a derived class that inherits from the base class, I want to transform the base instance into derived instance, (if possible without copying anything (maybe sending to the derived class a reference of the base class)) how can I achieve that?
Note: I need this because I'm using factory design pattern which identify the derived class needed to be created using a parameter located in the base instance.
//class A
//class B: public A (pure virtual)
//class C: public B
B BFactory::makeB(A &a) {
int n=a.getN();
if(n==1){
return new C();
}
}
Thanks.
Consider the case of the car.
You can treat a Lamborghini as a car.
You can treat a Yugo as a car.
You can treat a car as a Lamborghini if it is a Lamborghini. In C++ this means a pointer to car that really points to a Lamborghini. In order to get a Lamborghini pointer back out of the car pointer you should use dynamic_cast. If the car does not point to a Lamborghini, dynamic_cast will return NULL. This keeps you from trying to pass off a Yugo as a Lamborghini and blowing the Yugo's engine.
But when the Lamborghini is being treated as a car, it can only do car things. If you copy a Lamborghini into a car, you strip out all Lamborghini-ness forever. It's gone.
Code time!
This, I'm afraid cannot be done:
//class A
//class B: public A (pure virtual)
//class C: public B
B BFactory::makeB(A &a) {
int n=a.getN();
if(n==1){
return new C();
}
}
C is being copied into a B and the B is being returned. B would need a constructor that took a C, but the point is moot. B cannot be instantiated if it's pure virtual. For now we'll ignore the leak that would be new C()
Also can't use a reference for this job, pretty much the same problem, so you're trapped into returning a pointer
B * BFactory::makeB(A &a) {
int n=a.getN();
if(n==1){
return new C();
}
}
Now I'm going to make a suggestion: Build the make function into B and handle the case where A doesn't map to anything recognized by B.
class B: public A
{
public:
virtual ~B(){}
static B * makeB(A & a)
{
switch(a.getN())
{
case 1:
return new C();
}
return NULL;
}
};
But this leads to another recommendation: Why should B know anything? And What is the point of A at this level? Why is A storing build codes for classes two or more steps down the hierarchy? Bad from a maintenance point of view. The point of objects is they know who they are and how to manipulate themselves. Short-circuiting this leads to pain.
class B: public A
{
public:
virtual ~B(){}
virtual B* makeB() = 0;
};
Now B only makes Bs, needs no help from A, and those who extend B are stuck with figuring out how to make themselves--a task they should know better than anyone else. Much safer because there is never any possibility of a code unrecognised by B for a new class.
class C: public B
{
public:
B* makeB()
{
return new C();
}
};
class D: public B
{
public:
B* makeB()
{
return new D();
}
};
Edit: Traditional factory
You're asking for an abstract factory. For that you need nothing. You don't even need a class. You certainly don't need a class A. The goal of this sort of factory is the caller knows nothing about the class. By providing an A, the caller needs to know how to make an A or have another factory that makes an A.
First a bit of set-up in a header file BFactory.h:
#ifndef BFACTORY_H_
#define BFACTORY_H_
#include <exception>
class B
{
public:
virtual ~B(){}
virtual std::string whatAmI() = 0;
protected:
// data members common to all B subclasses
};
enum bType
{
gimmie_a_C,
gimmie_a_D,
gimmie_an_E
};
class BadTypeException: public std::exception
{
public:
const char* what() const noexcept
{
return "Dude! WTF?!?";
}
};
B* BFactory(enum bType type);
#endif /* BFACTORY_H_ */
Here I'm going to deviate from the book way a little. Rather than using an integer to identify the type to be built, I'm going to use an enum. Two reasons: Easier to read and understand gimme_a_C than 1 and generates a compiler error if you try to provide a value that is not enumerated.
enum bType
{
gimmie_a_C,
gimmie_a_D,
gimmie_an_E
};
And an exception to flag stupidity if the enum is updated with new types (gimmie_an_E) but the factory is not.
class BadTypeException: public std::exception
{
public:
const char* what() const noexcept
{
return "Dude! WTF?!?";
}
};
This is all the Factory client needs to see. They don't see C. They don't see D. They have no clue that C and D exist in any way other than the names listed in enum bType. All they ever see is pointers to B.
Now for the implementation BFactory.cpp:
#include "BFactory.h"
class C:public B
{
std::string whatAmI()
{
return "C";
}
};
class D:public B
{
std::string whatAmI()
{
return "D";
}
};
B* BFactory(enum bType type)
{
switch(type)
{
case gimmie_a_C:
return new C();
case gimmie_a_D:
return new C();
default:
throw BadTypeException();
}
}
I'll leave it up to the reader to spot the stupid bug in the above code that makes these error prone and why I don't like them.
And usage, main.cpp:
#include "BFactory.h"
int main()
{
B * temp;
temp = BFactory(gimmie_a_C);
std::cout << temp->whatAmI() << std::endl;
delete temp;
temp = BFactory(gimmie_a_D);
std::cout << temp->whatAmI() << std::endl;
delete temp;
//temp = BFactory(1001); // won't compile
try
{
temp = BFactory(gimmie_an_E); // will compile, throws exception
std::cout << temp->whatAmI() << std::endl;
}
catch(BadTypeException& wtf)
{
std::cerr << wtf.what() << std::endl;
}
}
There is still absolutely no use for or involvement of A. A if it exists, should no nothing about B or the children of B.
These days there is a little improvement we can make so that the pointers are a little safer. unique_ptr allows us to maintain the polymporphic advantages of a pointer to B without the memory management woes.
std::unique_ptr<B> BFactory(enum bType type)
{
switch(type)
{
case gimmie_a_C:
return std::unique_ptr<B>(new C());
case gimmie_a_D:
return std::unique_ptr<B>(new D());
default:
throw BadTypeException();
}
}
and the new main:
int main()
{
std::unique_ptr<B> temp;
temp = BFactory(gimmie_a_C);
std::cout << temp->whatAmI() << std::endl;
temp = BFactory(gimmie_a_D);
std::cout << temp->whatAmI() << std::endl;
}
You might want to define a constructor that takes the base class instance as the argument so you can later use static_cast to convert from the base class to the derived class.
class Derived : public Base
{
public:
Derived(const Base& base) : Base{base} {}
};
int main()
{
Base a;
Derived b = static_cast<Derived>(a);
}
If you want to create a derived class instance using the base class instance then there is some conversion rule between the two, which you can specify explicitly using a derived class constructor.
Although it is impossible to alter the type of an object you still can make instances of base and derived classes share the same data:
#include <memory>
#include <iostream>
class Base
{
protected:
struct CommonData
{
int A;
int B;
};
std::shared_ptr<CommonData> m_data;
public:
Base() : m_data(std::make_shared<CommonData>())
{
m_data->A = 0;
m_data->B = 0;
}
void SetData(Base * source)
{
m_data = source->m_data;
}
int A() const { return m_data->A; }
int B() const { return m_data->B; }
void SetA(int value) { m_data->A = value; }
void SetB(int value) { m_data->B = value; }
};
class Derived : public Base
{
public:
int C;
};
using namespace std;
int _tmain(int argc, _TCHAR* argv[])
{
Base base;
base.SetA(12);
base.SetB(46);
Derived derived;
derived.SetData(&base);
derived.C = 555;
cout << derived.A() << endl; // 12
cout << derived.C << endl; // 555;
cin.get();
}
A base class should not "know" about how to make its own derived class instances. That is the point of inheritance.
The "is a" relationship of derived classes means that any subclass instance will pass as a base class instance transparently, and you can treat it as one, and by default base class non-virtual methods are called on a base class reference, even if it a derived class instance. Only virtual methods use the derived class method.
In the case of creating a base class instance from a derived class you want to "slice" the instance data (normally a bad thing and normally a mistake).
class A{ // ... A stuff };
class B : A
{ // ... B stuff
A make_A() { return (A) B(*this); } // copy cast to A
};
Under no circumstances try to do this:
class B;
class A { // ...
B make_B() { return B(*this); }
};
That is inverted OO logic. It requires at least 2 scans of the source code, which C++ does not do. It fails.
I have a function that looks like:
// this function might modify base_ptr
void SomeFunction(shared_ptr<Base> &base_ptr)
{ if(some_condition) { base_ptr = some_other_ptr; } }
I'd like to call the function with a shared_ptr:
shared_ptr<Derived> d = ...;
SomeFunction(d);
This doesn't work though. It doesn't work if I'm using normal pointers either (ie. implicit casting to Base* & from Derived*. One workaround is to create a Base pointer from the Derived one, and then pass that to the function.
shared_ptr<Base> base = d;
SomeFunction(b);
But this isn't very pretty from a coding standpoint. It also adds confusion and the potential for a subtle bug:
shared_ptr<Derived> d = derived;
shared_ptr<Base> b = derived;
SomeFunction(b);
// b and d might now be pointing to different things -- not good!
Is there better way to do this?
What you are trying to do is inherently dangerous, and C++ is making it hard on purpose. Consider if C++ allowed you to call SomeFunction the way you wanted. Then you could do this:
struct Base {
};
struct Derived1 : Base {
void f1();
};
struct Derived2 : Base {
void f2();
};
void SomeFunction(shared_ptr<Base> &p)
{
p = make_shared<Derived2>(); // nothing wrong with converting
// a Derived2 pointer to a Base pointer.
}
int main()
{
shared_ptr<Derived1> d = make_shared<Derived1>();
SomeFunction(d); // An error, but what if it wasn't?
d->f1(); // Trying to call method f1 of a Derived2!
}
The compiler would not be able to know that d changed from a Derived1 pointer to a Derived2 pointer, so it wouldn't be able to give you a compile error when you tried to call method f1 of a Derived2.
You could template the function for the smart pointer's type
#include <iostream>
#include <memory>
#include <type_traits>
using namespace std;
class Base {
public:
virtual void hello() {
cout << "hello base" << endl;
}
};
class Derived : public Base {
public:
void hello() {
cout << "hello derived" << endl;
}
};
class otherClass {
public:
};
template<typename T>
void SomeFunction(shared_ptr<T> &base_ptr)
{
static_assert(is_base_of<Base, T>::value == true, "Wrong non-derived type");
base_ptr->hello();
// Rebase it
base_ptr.reset(new Derived);
base_ptr->hello();
}
int main() {
shared_ptr<Base> obj(new Base());
SomeFunction(obj);
// hello base
// hello derived
shared_ptr<Derived> obj2(new Derived());
// hello derived
// hello derived
SomeFunction(obj2);
shared_ptr<otherClass> obj3(new otherClass());
SomeFunction(obj3); // ASSERT
return 0;
}
http://ideone.com/ATqhEZ
If you intend to update all the smart pointers when you reset one, you'll have to do some book-keeping by yourself since they're not designed to have a "signal-like" mechanism to notify other smart pointers pointing to the same object.
Edited my answer to provide compile-time safety if you intend to use it with Base and relative subclasses.
Warning: the above solution uses C++11, the same could be accomplished in a similar way in pre-C++11 code
I am quite confused with the dynamic_cast keyword in C++.
struct A {
virtual void f() { }
};
struct B : public A { };
struct C { };
void f () {
A a;
B b;
A* ap = &b;
B* b1 = dynamic_cast<B*> (&a); // NULL, because 'a' is not a 'B'
B* b2 = dynamic_cast<B*> (ap); // 'b'
C* c = dynamic_cast<C*> (ap); // NULL.
A& ar = dynamic_cast<A&> (*ap); // Ok.
B& br = dynamic_cast<B&> (*ap); // Ok.
C& cr = dynamic_cast<C&> (*ap); // std::bad_cast
}
the definition says:
The dynamic_cast keyword casts a datum from one pointer or reference
type to another, performing a runtime check to ensure the validity of the cast
Can we write an equivalent of dynamic_cast of C++ in C so that I could better understand things?
Here's a rundown on static_cast<> and dynamic_cast<> specifically as they pertain to pointers. This is just a 101-level rundown, it does not cover all the intricacies.
static_cast< Type* >(ptr)
This takes the pointer in ptr and tries to safely cast it to a pointer of type Type*. This cast is done at compile time. It will only perform the cast if the types are related. If the types are not related, you will get a compiler error. For example:
class B {};
class D : public B {};
class X {};
int main()
{
D* d = new D;
B* b = static_cast<B*>(d); // this works
X* x = static_cast<X*>(d); // ERROR - Won't compile
return 0;
}
dynamic_cast< Type* >(ptr)
This again tries to take the pointer in ptr and safely cast it to a pointer of type Type*. But this cast is executed at runtime, not compile time. Because this is a run-time cast, it is useful especially when combined with polymorphic classes. In fact, in certain cases the classes must be polymorphic in order for the cast to be legal.
Casts can go in one of two directions: from base to derived (B2D) or from derived to base (D2B). It's simple enough to see how D2B casts would work at runtime. Either ptr was derived from Type or it wasn't. In the case of D2B dynamic_cast<>s, the rules are simple. You can try to cast anything to anything else, and if ptr was in fact derived from Type, you'll get a Type* pointer back from dynamic_cast. Otherwise, you'll get a NULL pointer.
But B2D casts are a little more complicated. Consider the following code:
#include <iostream>
using namespace std;
class Base
{
public:
virtual void DoIt() = 0; // pure virtual
virtual ~Base() {};
};
class Foo : public Base
{
public:
virtual void DoIt() { cout << "Foo"; };
void FooIt() { cout << "Fooing It..."; }
};
class Bar : public Base
{
public :
virtual void DoIt() { cout << "Bar"; }
void BarIt() { cout << "baring It..."; }
};
Base* CreateRandom()
{
if( (rand()%2) == 0 )
return new Foo;
else
return new Bar;
}
int main()
{
for( int n = 0; n < 10; ++n )
{
Base* base = CreateRandom();
base->DoIt();
Bar* bar = (Bar*)base;
bar->BarIt();
}
return 0;
}
main() can't tell what kind of object CreateRandom() will return, so the C-style cast Bar* bar = (Bar*)base; is decidedly not type-safe. How could you fix this? One way would be to add a function like bool AreYouABar() const = 0; to the base class and return true from Bar and false from Foo. But there is another way: use dynamic_cast<>:
int main()
{
for( int n = 0; n < 10; ++n )
{
Base* base = CreateRandom();
base->DoIt();
Bar* bar = dynamic_cast<Bar*>(base);
Foo* foo = dynamic_cast<Foo*>(base);
if( bar )
bar->BarIt();
if( foo )
foo->FooIt();
}
return 0;
}
The casts execute at runtime, and work by querying the object (no need to worry about how for now), asking it if it the type we're looking for. If it is, dynamic_cast<Type*> returns a pointer; otherwise it returns NULL.
In order for this base-to-derived casting to work using dynamic_cast<>, Base, Foo and Bar must be what the Standard calls polymorphic types. In order to be a polymorphic type, your class must have at least one virtual function. If your classes are not polymorphic types, the base-to-derived use of dynamic_cast will not compile. Example:
class Base {};
class Der : public Base {};
int main()
{
Base* base = new Der;
Der* der = dynamic_cast<Der*>(base); // ERROR - Won't compile
return 0;
}
Adding a virtual function to base, such as a virtual dtor, will make both Base and Der polymorphic types:
class Base
{
public:
virtual ~Base(){};
};
class Der : public Base {};
int main()
{
Base* base = new Der;
Der* der = dynamic_cast<Der*>(base); // OK
return 0;
}
Unless you're implementing your own hand-rolled RTTI (and bypassing the system one), it's not possible to implement dynamic_cast directly in C++ user-level code. dynamic_cast is very much tied into the C++ implementation's RTTI system.
But, to help you understand RTTI (and thus dynamic_cast) more, you should read up on the <typeinfo> header, and the typeid operator. This returns the type info corresponding to the object you have at hand, and you can inquire various (limited) things from these type info objects.
More than code in C, I think that an english definition could be enough:
Given a class Base of which there is a derived class Derived, dynamic_cast will convert a Base pointer to a Derived pointer if and only if the actual object pointed at is in fact a Derived object.
class Base { virtual ~Base() {} };
class Derived : public Base {};
class Derived2 : public Base {};
class ReDerived : public Derived {};
void test( Base & base )
{
dynamic_cast<Derived&>(base);
}
int main() {
Base b;
Derived d;
Derived2 d2;
ReDerived rd;
test( b ); // throw: b is not a Derived object
test( d ); // ok
test( d2 ); // throw: d2 is not a Derived object
test( rd ); // ok: rd is a ReDerived, and thus a derived object
}
In the example, the call to test binds different objects to a reference to Base. Internally the reference is downcasted to a reference to Derived in a typesafe way: the downcast will succeed only for those cases where the referenced object is indeed an instance of Derived.
First, to describe dynamic cast in C terms, we have to represent classes in C.
Classes with virtual functions use a "VTABLE" of pointers to the virtual functions.
Comments are C++. Feel free to reformat and fix compile errors...
// class A { public: int data; virtual int GetData(){return data;} };
typedef struct A { void**vtable; int data;} A;
int AGetData(A*this){ return this->data; }
void * Avtable[] = { (void*)AGetData };
A * newA() { A*res = malloc(sizeof(A)); res->vtable = Avtable; return res; }
// class B : public class A { public: int moredata; virtual int GetData(){return data+1;} }
typedef struct B { void**vtable; int data; int moredata; } B;
int BGetData(B*this){ return this->data + 1; }
void * Bvtable[] = { (void*)BGetData };
B * newB() { B*res = malloc(sizeof(B)); res->vtable = Bvtable; return res; }
// int temp = ptr->GetData();
int temp = ((int(*)())ptr->vtable[0])();
Then a dynamic cast is something like:
// A * ptr = new B();
A * ptr = (A*) newB();
// B * aB = dynamic_cast<B>(ptr);
B * aB = ( ptr->vtable == Bvtable ? (B*) aB : (B*) 0 );
The following is not really close to what you get from C++'s dynamic_cast in terms of type checking but maybe it will help you understand its purpose a little bit better:
struct Animal // Would be a base class in C++
{
enum Type { Dog, Cat };
Type type;
};
Animal * make_dog()
{
Animal * dog = new Animal;
dog->type = Animal::Dog;
return dog;
}
Animal * make_cat()
{
Animal * cat = new Animal;
cat->type = Animal::Cat;
return cat;
}
Animal * dyn_cast(AnimalType type, Animal * animal)
{
if(animal->type == type)
return animal;
return 0;
}
void bark(Animal * dog)
{
assert(dog->type == Animal::Dog);
// make "dog" bark
}
int main()
{
Animal * animal;
if(rand() % 2)
animal = make_dog();
else
animal = make_cat();
// At this point we have no idea what kind of animal we have
// so we use dyn_cast to see if it's a dog
if(dyn_cast(Animal::Dog, animal))
{
bark(animal); // we are sure the call is safe
}
delete animal;
}
A dynamic_cast performs a type checking using RTTI. If it fails it'll throw you an exception (if you gave it a reference) or NULL if you gave it a pointer.
There are no classes in C, so it's impossible to to write dynamic_cast in that language. C structures don't have methods (as a result, they don't have virtual methods), so there is nothing "dynamic" in it.
No, not easily. The compiler assigns a unique identity to every class, that information is referenced by every object instance, and that is what gets inspected at runtime to determine if a dynamic cast is legal. You could create a standard base class with this information and operators to do the runtime inspection on that base class, then any derived class would inform the base class of its place in the class hierarchy and any instances of those classes would be runtime-castable via your operations.
edit
Here's an implementation that demonstrates one technique. I'm not claiming the compiler uses anything like this, but I think it demonstrates the concepts:
class SafeCastableBase
{
public:
typedef long TypeID;
static TypeID s_nextTypeID;
static TypeID GetNextTypeID()
{
return s_nextTypeID++;
}
static TypeID GetTypeID()
{
return 0;
}
virtual bool CanCastTo(TypeID id)
{
if (GetTypeID() != id) { return false; }
return true;
}
template <class Target>
static Target *SafeCast(SafeCastableBase *pSource)
{
if (pSource->CanCastTo(Target::GetTypeID()))
{
return (Target*)pSource;
}
return NULL;
}
};
SafeCastableBase::TypeID SafeCastableBase::s_nextTypeID = 1;
class TypeIDInitializer
{
public:
TypeIDInitializer(SafeCastableBase::TypeID *pTypeID)
{
*pTypeID = SafeCastableBase::GetNextTypeID();
}
};
class ChildCastable : public SafeCastableBase
{
public:
static TypeID s_typeID;
static TypeID GetTypeID()
{
return s_typeID;
}
virtual bool CanCastTo(TypeID id)
{
if (GetTypeID() != id) { return SafeCastableBase::CanCastTo(id); }
return true;
}
};
SafeCastableBase::TypeID ChildCastable::s_typeID;
TypeIDInitializer ChildCastableInitializer(&ChildCastable::s_typeID);
class PeerChildCastable : public SafeCastableBase
{
public:
static TypeID s_typeID;
static TypeID GetTypeID()
{
return s_typeID;
}
virtual bool CanCastTo(TypeID id)
{
if (GetTypeID() != id) { return SafeCastableBase::CanCastTo(id); }
return true;
}
};
SafeCastableBase::TypeID PeerChildCastable::s_typeID;
TypeIDInitializer PeerChildCastableInitializer(&PeerChildCastable::s_typeID);
int _tmain(int argc, _TCHAR* argv[])
{
ChildCastable *pChild = new ChildCastable();
SafeCastableBase *pBase = new SafeCastableBase();
PeerChildCastable *pPeerChild = new PeerChildCastable();
ChildCastable *pSameChild = SafeCastableBase::SafeCast<ChildCastable>(pChild);
SafeCastableBase *pBaseToChild = SafeCastableBase::SafeCast<SafeCastableBase>(pChild);
ChildCastable *pNullDownCast = SafeCastableBase::SafeCast<ChildCastable>(pBase);
SafeCastableBase *pBaseToPeerChild = SafeCastableBase::SafeCast<SafeCastableBase>(pPeerChild);
ChildCastable *pNullCrossCast = SafeCastableBase::SafeCast<ChildCastable>(pPeerChild);
return 0;
}
static_cast< Type* >(ptr)
static_cast in C++ can be used in scenarios where all type casting can be verified at compile time.
dynamic_cast< Type* >(ptr)
dynamic_cast in C++ can be used to perform type safe down casting. dynamic_cast is run time polymorphism. The dynamic_cast operator, which safely converts from a pointer (or reference) to a base type to a pointer (or reference) to a derived type.
eg 1:
#include <iostream>
using namespace std;
class A
{
public:
virtual void f(){cout << "A::f()" << endl;}
};
class B : public A
{
public:
void f(){cout << "B::f()" << endl;}
};
int main()
{
A a;
B b;
a.f(); // A::f()
b.f(); // B::f()
A *pA = &a;
B *pB = &b;
pA->f(); // A::f()
pB->f(); // B::f()
pA = &b;
// pB = &a; // not allowed
pB = dynamic_cast<B*>(&a); // allowed but it returns NULL
return 0;
}
For more information click here
eg 2:
#include <iostream>
using namespace std;
class A {
public:
virtual void print()const {cout << " A\n";}
};
class B {
public:
virtual void print()const {cout << " B\n";}
};
class C: public A, public B {
public:
void print()const {cout << " C\n";}
};
int main()
{
A* a = new A;
B* b = new B;
C* c = new C;
a -> print(); b -> print(); c -> print();
b = dynamic_cast< B*>(a); //fails
if (b)
b -> print();
else
cout << "no B\n";
a = c;
a -> print(); //C prints
b = dynamic_cast< B*>(a); //succeeds
if (b)
b -> print();
else
cout << "no B\n";
}
dynamic_cast uses RTTI. It can slow down your application, you can use modification of the visitor design pattern to achieve downcasting without RTTI http://arturx64.github.io/programming-world/2016/02/06/lazy-visitor.html