RE: [Boost-users] How to get type given pointer to base?

30 Apr 2005


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-----Original Message-----
From: boost-users-bounces@lists.boost.org [mailto:boost-users-
bounces@lists.boost.org] On Behalf Of Stuart Dootson
Sent: Saturday, April 30, 2005 2:55 AM
To: boost-users@lists.boost.org
Subject: Re: [Boost-users] How to get type given pointer to base?
...
Hello all,
I hope somebody can help me with this. What I want to do is the
following:
Let's say I have a templated class called Foo
template<class U>
class Foo
{
public:
       Foo(U * p) : m_p(p) {}
       ~Foo() {}
       U* getP() const { return m_p; }
private:
       U* m_p;
};
Now let's say I want to have a collection of pointers to Foo. What
should I
do? The possibilities I can think of are the following:
1) Create a class FooBase and have Foo inherit from FooBase. FooBase is
non-templated so there should be no problem with having a collection of
pointers to FooBase where I can put instances of Foo. The problem with
...
approach is that given a FooBase* in that collection I could not use the
getP method because I would not know what type of Foo it holds.
2) Create a boost::variant of all the possible types of Foo<U>* I want
to
have. Then have a collection of this FooVariant. Using static visitation
I
can obtain the original type of Foo<U>* in a given FooVariant. The
On 4/30/05, Delfin Rojas  wrote:
this
problem
...
with this approach is that I get an ICE from VC++ 7.1 when the type of U
gets too complex (shared_ptr to another type for example).
Have I overlooked something? Is there a better technique for what I want
to
achieve? Any help would be great,
Thanks
Delfin Rojas
Using the first method, you could use something like this:
class FooBase
{
public:
   template<class T>
   T* getP() const { return
IsSameType(typeid(T*))?(T*)RawPointer():(T*)0; }
private:
   virtual bool IsSameType(std::type_info const& ti) const = 0;
   virtual void* RawPointer() const = 0;
};
template<class U>
class Foo : public FooBase
{
public:
   Foo(U * p) : m_p(p) {}
   ~Foo() {}
private:
   U* m_p;
   virtual bool IsSameType(std::type_info const& ti) const { return
0!=(ti == typeid(U*)); }
   virtual void* RawPointer() const { return (void*)m_p; }
};
In the following fragment
Foo<int> blah(new int);
int* pInt = blah.getP<int>();
   char* pChar = blah.getP<char>();
pInt is non-zero, pChar is zero because char != int.
Stuart Dootson
Yes, that would _work_ but unfortunately the code that need to call getP
would have to call it with every possible type and check if the return is
zero. Ideally I would like the following:

FooBase * pB = new Foo<int>(new int);
...
int * pI = pB->getP(); //compiles
char * pChar = pB->getP(); //does not compile

-delfin
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