From: "David B. Held" <dheld@codelogicconsulting.com>

"Joe Gottman" <jgottman@carolina.rr.com> wrote in message

...

template<class T> bool operator==(intrusive_ptr<T> const & a, T * b);

...

I'm curious why the pointer parameters are of type T* and not T const *.

Because that would not allow comparisons between types that are not T const*. Take T == int. With T*, you can compare int*. With T == int const, you get int const*. If you use T const* instead, you won't ever be able to compare int*.

int * p1(0); int const * p2(p1); // OK Notice also that T for instrusive_ptr<T> and for T * is one and the same, so you can't compare intrusive_ptr<T> with T const *. You could only compare intrusive_ptr<T const> with T const *. -- Rob Stewart stewart@sig.com Software Engineer http://www.sig.com Susquehanna International Group, LLP using std::disclaimer;

From: "David B. Held" <dheld@codelogicconsulting.com>

"Rob Stewart" <stewart@sig.com> wrote in message

...

[...] int * p1(0); int const * p2(p1); // OK

Notice also that T for instrusive_ptr<T> and for T * is one and the same, so you can't compare intrusive_ptr<T> with T const *. You could only compare intrusive_ptr<T const> with T const *.

And the problem with that is?

The original query was why the second arg wasn't T const *. My point is that comparing with a T * would work because T * can be converted to T const * implicitly. The other point is that if you have an instrusive_ptr<int const>, for example, you cannot compare it with an int *, because the function template expects the second argument to be of type int const *. (Actually, it will just complain about not being able to infer T since neither int nor int const will work.) -- Rob Stewart stewart@sig.com Software Engineer http://www.sig.com Susquehanna International Group, LLP using std::disclaimer;