Re: Templates: Template conversions

Carl Barron <>
Thu, 15 May 2008 00:04:43 CST
In article
Bharath <> wrote:

Hello All,
I was going thru section of Bjarne Stroustrup's The C++
Programming Language (3rd ed.).
Here I have a doubt on the first example of this section (copy pasted

template <class T> class Ptr
T* p ;
public :
      Ptr (T*);
      template <class T2> operator Ptr <T2> (); // convert Ptr<T> to
      // ...

I couldn't understand the below line in above example.
template <class T2> operator Ptr <T2> (); // convert Ptr<T> to Ptr<T2>

Can someone please explain what author is trying to achieve with above
line of code?
First of all, I couldn't understand after "operator" how a class name
can come (though this seems to be operator overloading ?)

Thanks in adv.

     first of all
     struct B{};
     struct A
          B b;
          operator B() {return b;}

      what operator does is provide a conversion from A to B;
      That is
       A x;
       B y;
       y = x; // converts z to a B and stores in y. without operator
B() above this is an illegal assignment of between different types.

      the templated example you quoted provides a conversion from
    Ptr<T1> to Ptr<T2> problably assuming a T1* is convertable to a T2*


    class A{};
    class B:A{};
    A *a;
    B *b;
    Ptr<A> pa(a)
    Ptr<B> pb(b);
    Ptr<A) pc(0);

    now since B derives from A a B * is an A *. Ptr<A> and Ptr<B> do not
provide this conversion, if template <class T2> operator Ptr<T1>() is
not provided, since it is

both pc = pa and pc = pb are allowed as would be the conversion of
plain pointers since B is derived from A.

if you have two user defined types A and B and you want implicit
conversion from A to B A needs an A::operator B();

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