Re: STL algorithms and output iterators

From:
"Victor Bazarov" <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Thu, 4 Oct 2007 23:19:53 -0400
Message-ID:
<fe4agp$95t$1@news.datemas.de>
Generic Usenet Account wrote:

I am trying to play with STL algorithms that have an output iterator
type as one of their arguments (copy, transform, replace_copy,
replace_copy_if. fill_n, generate_n, remove, recome_copy,
remove_copy_if, unique_copy, reverse_copy, rotate_copy, merge,
set_union, set_intersection, set_difference, set_symmetric_difference,
partial_sum, adjacent_difference).

All these algorithms operate on a "container" entity (e.g. list<int>
or set<string>). I am having a tough time getting a "wrapper"
function to compile if I pass both the container type and the
contained datatype as template arguments. For example, the compiler
likes <list<int> > but not <list, int>.

I am attaching some sample code to get my point across. Is there a
way to avoid the compilation error if I uncomment the line
#define BREAK_COMPILATION

Thanks,
Ramesh

///////////////////////////////////
//
// Sample program to test usage of STL algorithms
//
///////////////////////////////////

#include <iostream>
#include <iterator>
#include <string>
#include <list>
#include <set>

#include <algorithm>

using namespace std;

//#define BREAK_COMPILATION


The rest of the code presumes we uncommented this. I redacted
unnecessary code

template<typename T1, typename T2>
void
executeSTLAlgo(const string& label)
{

[..]

}

main()


int main()

{
 executeSTLAlgo<list,int>("List");
 executeSTLAlgo<set, int>("Set");
}


That's not going to work because 'list' is NOT a typename. It's
a template. In order to use 'list' as is (or 'std::list') you need
to declared your 'executeSTLAlgo' to have a _template_ template
argument (no, I didn't make a mistake, two "templates"). Find
a good book ("C++ Templates" by Vandevoorde and Josuttis is good)
and read about template template arguments.

V
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