Question

It seems like the compiler is very close to doing what I want (because it calls out my function as a candidate), but I have no idea what I’m doing wrong.

#include <stdio.h>
#include <stdlib.h>
#include <list>

using namespace std;

template <class U, template<class U> class T>
void AppendSorted( T<U>& l, U val )
{
  typename T<U>::reverse_iterator rt = l.rbegin();

  while( ((*rt) > val) && (rt != l.rend()) )
    rt++;

  l.insert( rt.base(), val );
}

int main( int argc, char* argv[] )
{
    list<int> foo;
    AppendSorted<int, list<int> >( foo, 5 );

    list<int>::iterator i;
    for( i = foo.begin(); i != foo.end(); i++ )
    {
        printf("%d\n",*i);
    }

    return 0;
}

The error I’m getting is:

test.cpp: In function ‘int main(int, char**)’:
test.cpp:21:43: error: no matching function for call to ‘AppendSorted(std::list<int>&, int)’
test.cpp:21:43: note: candidate is:
test.cpp:8:6: note: template<class U, template<class U> class T> void AppendSorted(T<U>&, U)

Answer 1

This signature

template <class U, template<class U> class T>
void AppendSorted( T<U>& l, U val )

indicates that there will be one concrete type (class U) and one template (template<class U> class T).

This invocation

AppendSorted<int, list<int> >( foo, 5 );

provides two concrete types, int and list<int>. list is a template, list<int> is a concrete type, a template instance, but not a template.

Just change the function to accept the concrete type of the collection:

template <class U, class T>
void AppendSorted( T& l, U val )
{
  typename T::reverse_iterator /* or auto */ rt = l.rbegin();

  while( (rt != l.rend()) && ((*rt) > val) )
    rt++;

  l.insert( rt.base(), val );
}

And let the compiler infer the type arguments, instead of specifying them.

AppendSorted( foo, 5 );