问题
I was thinking along the lines of using typeid() but I don\'t know how to ask if that type is a subclass of another class (which, by the way, is abstract)
回答1:
You really shouldn't. If your program needs to know what class an object is, that usually indicates a design flaw. See if you can get the behavior you want using virtual functions. Also, more information about what you are trying to do would help.
I am assuming you have a situation like this:
class Base;
class A : public Base {...};
class B : public Base {...};
void foo(Base *p)
{
if(/* p is A */) /* do X */
else /* do Y */
}
If this is what you have, then try to do something like this:
class Base
{
virtual void bar() = 0;
};
class A : public Base
{
void bar() {/* do X */}
};
class B : public Base
{
void bar() {/* do Y */}
};
void foo(Base *p)
{
p->bar();
}
Edit: Since the debate about this answer still goes on after so many years, I thought I should throw in some references. If you have a pointer or reference to a base class, and your code needs to know the derived class of the object, then it violates Liskov substitution principle. Uncle Bob calls this an "anathema to Object Oriented Design".
回答2:
class Base
{
public: virtual ~Base() {}
};
class D1: public Base {};
class D2: public Base {};
int main(int argc,char* argv[]);
{
D1 d1;
D2 d2;
Base* x = (argc > 2)?&d1:&d2;
if (dynamic_cast<D2*>(x) == nullptr)
{
std::cout << "NOT A D2" << std::endl;
}
if (dynamic_cast<D1*>(x) == nullptr)
{
std::cout << "NOT A D1" << std::endl;
}
}
回答3:
You can do it with dynamic_cast (at least for polymorphic types).
Actually, on second thought--you can't tell if it is SPECIFICALLY a particular type with dynamic_cast--but you can tell if it is that type or any subclass thereof.
template <class DstType, class SrcType>
bool IsType(const SrcType* src)
{
return dynamic_cast<const DstType*>(src) != nullptr;
}
回答4:
dynamic_cast can determine if the type contains the target type anywhere in the inheritance hierarchy (yes, it's a little-known feature that if B inherits from A and C, it can turn an A* directly into a C*). typeid() can determine the exact type of the object. However, these should both be used extremely sparingly. As has been mentioned already, you should always be avoiding dynamic type identification, because it indicates a design flaw. (also, if you know the object is for sure of the target type, you can do a downcast with a static_cast. Boost offers a polymorphic_downcast that will do a downcast with dynamic_cast and assert in debug mode, and in release mode it will just use a static_cast).
回答5:
I disagree that you should never want to check an object's type in C++. If you can avoid it, I agree that you should. Saying you should NEVER do this under any circumstance is going too far though. You can do this in a great many languages, and it can make your life a lot easier. Howard Pinsley, for instance, showed us how in his post on C#.
I do a lot of work with the Qt Framework. In general, I model what I do after the way they do things (at least when working in their framework). The QObject class is the base class of all Qt objects. That class has the functions isWidgetType() and isWindowType() as a quick subclass check. So why not be able to check your own derived classes, which is comparable in it's nature? Here is a QObject spin off of some of these other posts:
class MyQObject : public QObject
{
public:
MyQObject( QObject *parent = 0 ) : QObject( parent ){}
~MyQObject(){}
static bool isThisType( const QObject *qObj )
{ return ( dynamic_cast<const MyQObject*>(qObj) != NULL ); }
};
And then when you are passing around a pointer to a QObject, you can check if it points to your derived class by calling the static member function:
if( MyQObject::isThisType( qObjPtr ) ) qDebug() << "This is a MyQObject!";
回答6:
I don't know if I understand your problem correctly, so let me restate it in my own words...
Problem: Given classes B and D, determine if D is a subclass of B (or vice-versa?)
Solution: Use some template magic! Okay, seriously you need to take a look at LOKI, an excellent template meta-programming library produced by the fabled C++ author Andrei Alexandrescu.
More specifically, download LOKI and include header TypeManip.h from it in your source code then use the SuperSubclass class template as follows:
if(SuperSubClass<B,D>::value)
{
...
}
According to documentation, SuperSubClass<B,D>::value will be true if B is a public base of D, or if B and D are aliases of the same type.
i.e. either D is a subclass of B or D is the same as B.
I hope this helps.
edit:
Please note the evaluation of SuperSubClass<B,D>::value happens at compile time unlike some methods which use dynamic_cast, hence there is no penalty for using this system at runtime.
回答7:
#include <stdio.h>
#include <iostream.h>
class Base
{
public: virtual ~Base() {}
template<typename T>
bool isA() {
return (dynamic_cast<T*>(this) != NULL);
}
};
class D1: public Base {};
class D2: public Base {};
class D22: public D2 {};
int main(int argc,char* argv[]);
{
D1* d1 = new D1();
D2* d2 = new D2();
D22* d22 = new D22();
Base* x = d22;
if( x->isA<D22>() )
{
std::cout << "IS A D22" << std::endl;
}
if( x->isA<D2>() )
{
std::cout << "IS A D2" << std::endl;
}
if( x->isA<D1>() )
{
std::cout << "IS A D1" << std::endl;
}
if(x->isA<Base>() )
{
std::cout << "IS A Base" << std::endl;
}
}
Result:
IS A D22
IS A D2
IS A Base
回答8:
You can only do it at compile time using templates, unless you use RTTI.
It lets you use the typeid function which will yield a pointer to a type_info structure which contains information about the type.
Read up on it at Wikipedia
回答9:
In c# you can simply say:
if (myObj is Car) {
}
回答10:
The code below demonstrates 3 different ways of doing it:
- virtual function
- typeid
- dynamic_cast
#include <iostream>
#include <typeinfo>
#include <typeindex>
enum class Type {Base, A, B};
class Base {
public:
virtual ~Base() = default;
virtual Type type() const {
return Type::Base;
}
};
class A : public Base {
Type type() const override {
return Type::A;
}
};
class B : public Base {
Type type() const override {
return Type::B;
}
};
int main()
{
const char *typemsg;
A a;
B b;
Base *base = &a; // = &b; !!!!!!!!!!!!!!!!!
Base &bbb = *base;
// below you can replace base with &bbb and get the same results
// USING virtual function
// ======================
// classes need to be in your control
switch(base->type()) {
case Type::A:
typemsg = "type A";
break;
case Type::B:
typemsg = "type B";
break;
default:
typemsg = "unknown";
}
std::cout << typemsg << std::endl;
// USING typeid
// ======================
// needs RTTI. under gcc, avoid -fno-rtti
std::type_index ti(typeid(*base));
if (ti == std::type_index(typeid(A))) {
typemsg = "type A";
} else if (ti == std::type_index(typeid(B))) {
typemsg = "type B";
} else {
typemsg = "unknown";
}
std::cout << typemsg << std::endl;
// USING dynamic_cast
// ======================
// needs RTTI. under gcc, avoid -fno-rtti
if (dynamic_cast</*const*/ A*>(base)) {
typemsg = "type A";
} else if (dynamic_cast</*const*/ B*>(base)) {
typemsg = "type B";
} else {
typemsg = "unknown";
}
std::cout << typemsg << std::endl;
}
The program above prints this:
type A
type A
type A
回答11:
I was thinking along the lines of using
typeid()...
Well, yes, it could be done by comparing: typeid().name(). If we take the already described situation, where:
class Base;
class A : public Base {...};
class B : public Base {...};
void foo(Base *p)
{
if(/* p is A */) /* do X */
else /* do Y */
}
A possible implementation of foo(Base *p) would be:
#include <typeinfo>
void foo(Base *p)
{
if(typeid(*p) == typeid(A))
{
// the pointer is pointing to the derived class A
}
else if (typeid(*p).name() == typeid(B).name())
{
// the pointer is pointing to the derived class B
}
}
来源：https://stackoverflow.com/questions/307765/how-do-i-check-if-an-objects-type-is-a-particular-subclass-in-c