Forward declaring standard library classes

[Jonathan Wakely]

On Thu, Nov 18, 2004 at 10:05:34PM +0000, Chris Jefferson wrote:

> However, I don't want to pull in <vector>, <list>, <deque>, <map>, 
> etc... so I forward declare these classes. However, the following 
> doesn't compile:
> 
> ***
> #include<functional>
> 
> namespace std
> {
>  template<typename _Tp>
>  class allocator;
>  template<class _Key, class _Compare = less<_Key>,
>           class _Alloc = allocator<_Key> >
>  class set;
> }
> 
> #include<set>
> ***
> 
> complaining that I have redefined the default parameter to _Compare. It 
> (unsuprisingly) doesn't work with my original version, where I also 
> forward declare less.
> 
> At the moment I'm getting around this by simply including every standard 
> container in stl_algo.h, but obviously this cannot continue. Any 
> suggestions?

If you forward declare the templates then the default parameters must
only appear on the initial forward declaration. Look at how it's done
with bits/stringfwd.h and bits/basic_string.h

You could have a vectorfwd.h header that can be used by stl_algo.h and
is included at the top of vector. This would make the name "std::vector"
visible (but not usable) when <algorithm> is included (and similarly for
all other types you do this for), which may not be acceptable. 


How about having an stl_algo_hasswap.h header that contains just:

 template<typename _Tp>
    struct _UseSwap
    { enum{_M_type = 0};};

which is enough for it to be used in stl_algo.h

Then in vector include stl_algo_hasswap.h and add a specialisation:

  template<typename _Tp>
    struct _UseSwap<std::vector<_Tp> >
    {enum{_M_type = 1};};

and similarly in map. include stl_algo_hasswap.h and specialise:

  template<typename _Tp>
    struct _UseSwap<std::map<_Tp> >
    {enum{_M_type = 1};};

etc. etc.

It's still early and I've only just had some coffee, but that should work,
because before anyone can use an algorithm on a vector they have to have
included <vector>, at which point the specialisation is visible and will
be used.

jon

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