Hello!

I have just discovered a way to use the private implementation idiom
(pimpl), without the overhead of dynamic memory allocation. For those of
you who don't know what this is, Wikipedia has a nice article you can
read. Anyway, I discovered that if you make all members in the
implementation class mutable, you can in fact use this idiom without any
"unnecessary" memory allocation. Here's a minimal example of the method:

// In the header of your class called Line

#include <string>

class Line
{
public:

Line(const std::string& name);
const std::string& GetName() const;
void SetName(const std::string& s);

private:

// Private implementation idiom:
// all member variables are hidden in this class
class LineImpl;
const LineImpl& m_pimpl; // normally a non-const pointer
};

// and in your implementation file:

#include "Line.h"

// Here we define the class with the member variables
class Line::LineImpl
{
public:

LineImpl(const std::string& s) : m_s(s) {}
// all methods need to be const here
const std::string& GetName() const { return m_s; }
void SetName(const std::string& s) const { m_s = s; }

private:

mutable std::string m_s; // the trick! all members are mutable
};

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

// forward all member functions to the private implementation
const std::string& Line::GetName() const
{
return m_pimpl.GetName();
}

void Line::SetName(const std::string& s)
{
m_pimpl.SetName(s);
}

Ok experts, what do you all think? This method sacrifies
const-correctness for some extra speed. Is it worth it?

--
Daniel

Daniel Lidström a écrit :
> Hello!
>
> I have just discovered a way to use the private implementation idiom
> (pimpl), without the overhead of dynamic memory allocation. For those of
> you who don't know what this is, Wikipedia has a nice article you can
> read. Anyway, I discovered that if you make all members in the
> implementation class mutable, you can in fact use this idiom without any
> "unnecessary" memory allocation. Here's a minimal example of the method:
>
> // In the header of your class called Line
>
> #include <string>
>
> class Line
> {
> public:
>
> Line(const std::string& name);
> const std::string& GetName() const;
> void SetName(const std::string& s);
>
> private:
>
> // Private implementation idiom:
> // all member variables are hidden in this class
> class LineImpl;
> const LineImpl& m_pimpl; // normally a non-const pointer
> };
>
> // and in your implementation file:
>
> #include "Line.h"
>
> // Here we define the class with the member variables
[snip]
> // all methods need to be const here
[snip]
> mutable std::string m_s; // the trick! all members are mutable

Which mean you coerce the code into compilation. That's all.

> // 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 ?

> [snip]
> Ok experts, what do you all think? This method sacrifies
> const-correctness for some extra speed. Is it worth it?

Not really.
And certainly not worth a dangling reference.


Michael

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

> Daniel Lidström a écrit :
> > // 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.

--
Daniel

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

Michael

Daniel Lidström <somebody@microsoft.com> wrote:

> // In the header of your class called Line
>
> #include <string>
>
> class Line
> {
> public:
>
> Line(const std::string& name);
> const std::string& GetName() const;
> void SetName(const std::string& s);
>
> private:
>
> // Private implementation idiom:
> // all member variables are hidden in this class
> class LineImpl;
> const LineImpl& m_pimpl; // normally a non-const pointer
> };
>
> // and in your implementation file:
>
> #include "Line.h"
>
> // Here we define the class with the member variables
> class Line::LineImpl
> {
> public:
>
> LineImpl(const std::string& s) : m_s(s) {}
> // all methods need to be const here
> const std::string& GetName() const { return m_s; }
> void SetName(const std::string& s) const { m_s = s; }
>
> private:
>
> mutable std::string m_s; // the trick! all members are mutable
> };
>
> // create the pimpl instance without using new
> Line::Line(const std::string& s) : m_pimpl(LineImpl(s)) {}

Where would the memory for the LineImpl object be placed? It isn't
embedded in the object, nor is it in the heap, and it can't be placed on
the stack (and still survive the call to the c_tor.)

Doesn't sound like a good idea to me.

Daniel Lidström wrote:
> I have just discovered a way to use the private implementation idiom
> (pimpl), without the overhead of dynamic memory allocation.
[ storing a reference to a temporary ]

As you noticed, this doesn't work. However, there is a method that works.
All you have to do is to add a suitably aligned and sufficiently large
buffer into the class:

class foo {
aligned_storage<42> m_impl;
class implementation;
foo();
~foo();
void some_function();
};

class foo::implementation { ... };

foo::foo() {
// placement new
new m_impl.get<void>() implementation;
}
foo::~foo() {
// explicit dtor invokation
m_impl.get<implementation>()->~implementation;
}
void foo::some_function() {
m_impl.get<implementation>()->some_function();
}

Is it worth the hassle? Typically not, in particular since it's hard to
guarantee that you have both enough but still not too much memory.

Uli

On Oct 17, 10:36 pm, "Daniel T." <danie...@earthlink.net> wrote:
> Daniel Lidström <someb...@microsoft.com> wrote:
> > // In the header of your class called Line

> > #include <string>

> > class Line
> > {
> > public:
> > Line(const std::string& name);
> > const std::string& GetName() const;
> > void SetName(const std::string& s);

> > private:
> > // Private implementation idiom:
> > // all member variables are hidden in this class
> > class LineImpl;
> > const LineImpl& m_pimpl; // normally a non-const pointer
> > };

> > // and in your implementation file:

> > #include "Line.h"

> > // Here we define the class with the member variables
> > class Line::LineImpl
> > {
> > public:
> > LineImpl(const std::string& s) : m_s(s) {}
> > // all methods need to be const here
> > const std::string& GetName() const { return m_s; }
> > void SetName(const std::string& s) const { m_s = s; }
> > private:
> > mutable std::string m_s; // the trick! all members are mutable
> > };

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

> Where would the memory for the LineImpl object be placed? It
> isn't embedded in the object, nor is it in the heap, and it
> can't be placed on the stack (and still survive the call to
> the c_tor.)

>From the standard (§12.2/5): "A temporary bound to a reference
member in a constructor's ctor-initializer persists until the
constructor exits." In colloquial terms: the temporary is
created on the stack, and destructed before returning from the
constructor.

> Doesn't sound like a good idea to me.

It isn't, unless you like undefined behavior and hard to find
bugs.

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

In article <47165e09$0$26728$426a74cc@news.free.fr>,
Michael DOUBEZ <michael.doubez@free.fr> wrote:
>Daniel Lidström a écrit :
>> In article <4716585e$0$25087$426a74cc@news.free.fr>,
>> Michael DOUBEZ <michael.doubez@free.fr> wrote:
>>
>>> Daniel Lidström a écrit :
>>>> // 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();
....