Re: fwd declaring STL containers

From:

Mark P <usenet@fall2005REMOVE.fastmailCAPS.fm>

Newsgroups:

alt.comp.lang.learn.c-c++,comp.lang.c++

Date:

Wed, 21 Jun 2006 05:43:40 GMT

Message-ID:

<g05mg.54698$Lm5.4870@newssvr12.news.prodigy.com>

LR wrote:

Mark P wrote:

LR wrote:

Mark P wrote:

Is there any way to forward declare STL container classes such as
list, set, map, etc.? (My impression is that there isn't, since
these are all defined in std.)

Failing that, consider the following snippet of code:

//////////

#include <list>

template <class Ty = int>
struct Foo
{
typedef std::list<Ty> Type;
};

//////////

If this block of code were included in a translation unit that never
made any further reference to Foo or Foo::Type, is it reasonable to
assume that the compiled code would not be any larger? (I
understand this is an implementation issue, but your experience and
intuition would be very helpful.) FWIW, my testing on gcc indicates
no difference.

Could you expand on this a little bit?

Have you tried to compare something like:

int main() {
static Foo f;
}

and

int main() { }

My intuition tells me these will be different sizes. I tried with
two compilers, with the first, the object file size changed, but not
the executable file size. With the second, both files changed size.

Did you mean the executable file size? Object file size? Footprint
in memory at runtime?

I looked at object file size and executable file size and saw no
difference. I don't know that your example is particularly relevant
to my issue though. I never instantiate my Foo object-- it's only
used to emulate a templated typedef.

I was only looking at what you said here: "If this block of code were
included in a translation unit that never made any further reference to
Foo" But probably I misinterpreted what you meant.

But if you never instantiate, yet you have some need to wrap the typedef
into a scope of some kind, have you considered a namespace?

The point is not simply to wrap the typedef inside of a scope, but
rather to obtain something like a templated typedef since the language
has no "direct" support for such a thing.

[If you're curious, I have a bunch of these wrapped typedefs for
various STL container classes which I use to supply my own default
allocator. This in turn simplifies the client syntax significantly.
However, they're all stuck together in a single header file which
includes many of the STL container headers, even though any
particular user of the header may only need some of them.]

Now I'm curious. How does this simplify client syntax?

Compare the following two declarations:

std::map<Key, Ty, std::less<Key>,
myAlloc<std::pair<const Key,Ty> > > myMap;

my_map<Key,Ty>::Type myMap;

The issue is that the allocator parameter is the last among all
parameters so to override the default it's necessary to specify all
parameters. Compound this with the particularly unwieldy value_type
of the map, and it gets pretty ugly.

-Mark

Even more curious, why do you need to specify an allocator?

Just the nature of the project I'm working on. All memory allocations
are handled by a memory manager. My allocator serves as an interface
between STL allocation conventions and the memory manager.