7f57d9b5   
   From: daniel.kruegler@googlemail.com   
      
   On 2011-08-25 01:44, Iker Jamardo Zugaza wrote:   
   > I have looked over the internet and this thread looking for a complete   
   > answer of this situation I am facing. I have read that throwing smart   
   > pointers to objects is not very clever. I just want to understand why   
   > is this happening. I will explain the situation. Let's imagine this   
   > simple hierarchy:   
   >   
   > class Foo   
   > {   
   > public: virtual ~Foo() {}   
   > };   
   >   
   > typedef tr1::shared_ptr  SPFoo;   
   >   
   > class FooInherited: public Foo { };   
   >   
   > typedef tr1::shared_ptr  SPFooInherited;   
      
   Let me alert you here, that even though FooInherited is derived from   
   Foo, there is no such inheritance relation betweeb SPFoo and SPFooInherited.   
      
   > And let's check this test code:   
   >   
   > int main(int argc, char** argv)   
   > {   
   >    try   
   >    {   
   >      throw FooInherited();   
   >    }   
   >    catch(const Foo&  f)   
   >    {   
   >      cout<<  "Foo&  caught!"<<  endl;   
   >    }   
      
   This works, because FooInherited is a sub-class of Foo. The excpetion   
   object has type FooInherited and can thus enter a handler for type Foo.   
      
   >    try   
   >    {   
   >      throw SPFooInherited(new FooInherited());   
   >    }   
   >    catch(const SPFoo&  f)   
   >    {   
   >      cout<<  "SPFoo&  caught!"<<  endl;   
   >    }   
   >   
   >    return 0;   
   > }   
   >   
   > Everythign compiles but in runtime the second try-catch won't be   
   > executed. Can someone explain me why? Specially if lines of code like   
   > this work perfectly fine in runtime.   
      
   The incorrect assumption of this code is that an exception handler for   
   SPFoo could be entered with an exception object of type SPFooInherited.   
   This is not the case, because there does not exist any type-inheritance   
   relation between SPFooInherited and SPFoo. In fact, both SPFooInherited   
   and SPFoo are completely unrelated types.   
      
   HTH & Greetings from Bremen,   
      
   Daniel Krügler   
      
      
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