Le Tue, 31 Jan 2006 11:06:51 -0500, me22 a écrit :
On 31/01/06, Pierre THIERRY <nowhere.man@levallois.eu.org> wrote:
+ shared_ptr & operator=(T * r) + { + this_type(r).swap(*this); + return *this; + } +
That results in implicitly creating a shared_ptr from a raw pointer, which as I understand it is explicitly disallowed because it means that simple code such as the following will fail: int *p = new int; shared_ptr<int> sp; sp = p; sp = p;
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);
My understanding is that the operator=(T*) is purposefully not there to make this more explicit.
But this also renders some uses impossible, like this one: stack<foo*> fooS; while(shared_ptr<foo> = fooS.pop()) { ... } Sorry for answering so late to this (now) old thread. If I should start it again for clarity, just tell me. Lately, Nowhere man -- nowhere.man@levallois.eu.org OpenPGP 0xD9D50D8A