Re: Pimpl idiom without dynamic memory allocation

From:

Kai-Uwe Bux <jkherciueh@gmx.net>

Newsgroups:

comp.lang.c++

Date:

Thu, 18 Oct 2007 07:35:35 -0700

Message-ID:

<ff7r0f$rg6$1@murdoch.acc.Virginia.EDU>

Yannick Tremblay wrote:

In article <47165e09$0$26728$426a74cc@news.free.fr>,
Michael DOUBEZ <michael.doubez@free.fr> wrote:

Daniel Lidstr?m a ?crit :

In article <4716585e$0$25087$426a74cc@news.free.fr>,
Michael DOUBEZ <michael.doubez@free.fr> wrote:

Daniel Lidstr?m a ?crit :

// create the pimpl instance without using new
Line::Line(const std::string& s) : m_pimpl(LineImpl(s)) {}

Your local is destroyed when going out of scope. Doesn't it ?

No it isn't. It is actually ok to bind a temporary object to a const
reference. There will be no "dangling" reference.

It is ok to bind it but that doesn't mean the lifetime of the temporary
is extended.

Example:
const std::string& foo()
{
return std::string("bar");
}

The value returned by foo() is an dangling reference.

Euh, it's not what he is doing, it's more like:

std::string foo()
{
return std::string("bar");
}

int main()
{
std::string const & val = foo();
...

It's neither. What he is doing is initialzing a reference member from a
temporary object. The lifetime of that object lasts exactly to the end of
the constructor call. Afterwards, i.e., for the entire lifetime of the
fully constructed object, the reference is dangling. From the standard:

  [...] A temporary bound to a reference member in a constructor?s
  ctor-initializer (12.6.2) persists until the constructor exits. ...
  [12.2/5]

This provision makes you wonder. What is the point of restricting the
life-time of the temporary to the duration of the constructor if the object
thus initialized is bound to have a dangling reference ever after?

Best

Kai-Uwe Bux