On 6/12/06, Pierre THIERRY <nowhere.man@levallois.eu.org> wrote:
I understand, but the following code would fail the same way without compilation warnings:
int p* = new int; shared_ptr<int> sp; sp.reset(p); sp.reset(p);
But in that case you're typing reset, which means you're aware that you're using smart pointers. You also would never have "sp.reset(p);" if sp used to be a plain pointer, so it's not a problem when refitting existing code to use smart pointers.
But this also renders some uses impossible, like this one:
stack<foo*> fooS; while(shared_ptr<foo> = fooS.pop()) { ... }
I'm not sure what that code's trying to do. Each of these works fine: stack<foo*> fooS; shared_ptr<foo> fptr; while ( fptr.reset(fooS.pop()) ) { ... } stack<foo*> fooS; shared_ptr<foo> fptr; while ( fptr = shared_ptr<foo>(fooS.pop()) ) { ... } stack<foo*> fooS; shared_ptr<foo> fptr; template <typename T> shared_ptr<T> make_shared_ptr(T *ptr) { return shared_ptr<foo>(ptr); } while ( fptr = make_shared_ptr(fooS.pop()) ) { ... } stack<foo*> fooS; if ( shared_ptr<foo> fptr( fooS.pop() ) ) { ... }
Sorry for answering so late to this (now) old thread. If I should start it again for clarity, just tell me.
Luckily gmail still managed to thread it for me =) HTH, ~ Scott McMurray