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 01:11:23 GMT
Message-ID:
<%01mg.69823$4L1.39123@newssvr11.news.prodigy.com>
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.

[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

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