Re: Construct a string with char* taking ownership

From:

"Nevin :-] Liber" <nevin@eviloverlord.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Tue, 4 Nov 2008 11:18:15 CST

Message-ID:

<nevin-68227C.15182803112008@chi.news.speakeasy.net>

In article
<d0c97afd-ddbd-4a62-b343-a50b7d1cfa22@w39g2000prb.googlegroups.com>,
cbarron413@roadrunner.com wrote:

As you wrote the code free is Ptf and any Ptf could be substituted
later since its type is public. [Its a plain old function ptr]

How does it get substituted? The member variable in question is private:

struct shared_c_string
{
     shared_c_string(char* s = 0) : ptr(s, free) {}
     operator const char*() const { return ptr.get(); }

private:
     boost::shared_ptr<char> ptr;
};

private:
struct deleter {void operator () (void *p) {free(p);}

and use a default constructed deleter in the ctor of the shared_ptr
the
type is not publicly available so the deleter can't be changed by s
user.

Does this make sense???

I'm still not getting it. If the variable isn't publicly accessible,
why does it matter what the underlying type of its internal deleter is?

If you could post some example code which demonstrates the problem
(changes the deleter for the private ptr member variable, unless I'm
misunderstanding what the actual problem is) it would aid greatly in
understanding the issue.

--
  Nevin ":-)" Liber <mailto:nevin@eviloverlord.com> 773 961-1620

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