Re: [boost] explicit shared_ptr(weak_ptr<Y> const & r);

From: John Torjo (john.lists_at_[hidden])
Date: 2004-10-18 02:53:19

> 
> You can't accidentally omit the lock(), and shared_ptr<T> p( weak ) is 
> invalid in an if statement.
> 
> shared_ptr<T> val( weak ); does exactly what it is intended to do.
> 

It does what it is documented to do. But I really fail to see any code 
that would actually prefer this:

shared_ptr<T> val( weak );

over this:

shared_ptr<T> val = weak.lock();

Thus, people might use the former synax by mistake. At least, I did it ;)

Basically, actually my mistake was more subtle:

struct wnd {
    shared_ptr<...> p;
    // bug here: it should have actually been shared_ptr<...>
    wnd( weak_ptr<...> p) : p(p) {}
};

struct weak_wnd {
     weak_ptr<...> p;
     shared_ptr<...> to_wnd() const { return p.lock(); }
};

weak_wnd a;
// a.to_wnd():
// a shared pointer gets constructed, which is automatically converted
// to a weak_ptr, which then is converted to a shared_ptr
wnd b = a.to_wnd();

Had that constuctor not been public, this would have generated a 
compile-time error.

Best,
John

-- 
John Torjo,    Contributing editor, C/C++ Users Journal
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