C++ syntax issue

It's late and I can't figu开发者_如何学Cre out what is wrong with my syntax. I have asked other people and they can't find the syntax error either so I came here on a friend's advice.

template <typename TT>
bool PuzzleSolver<TT>::solve ( const Clock &pz ) {

  possibConfigs_.push( pz.getInitial() );
  vector< Configuration<TT> > next_;

  //error is on next line
  map< Configuration<TT> ,Configuration<TT> >::iterator found;

  while ( !possibConfigs_.empty() && possibConfigs_.front() != pz.getGoal() ) {
    Configuration<TT> cfg = possibConfigs_.front();
    possibConfigs_.pop();
    next_ = pz.getNext( cfg );

    for ( int i = 0; i < next_.size(); i++ ) {
      found = seenConfigs_.find( next_[i] );
      if ( found != seenConfigs_.end() ) {
        possibConfigs_.push( next_[i] );
        seenConfigs_.insert( make_pair( next_[i], cfg ) );
      }
    }
  }
}

What is wrong?

Thanks for any help.

If I remember correctly, this syntax is ambiguous:

map< Configuration<TT> ,Configuration<TT> >::iterator found;

Try that instead:

typename map< Configuration<TT> ,Configuration<TT> >::iterator found;

For dependent names you need to use the typename keyword. A dependent name is a name that depends on a template parameter.

So you have to:

typename map< Configuration<TT> ,Configuration<TT> >::iterator found;

This is necessary because the type of your map isn't known until your class (PuzzleSolver) is instantiated, since it depends on the template parameter TT.

This is a classic example of the required use of typename to identify named types that depend on the template type parameter. You want to use

typename map< Configuration<TT>, Configuration<TT> >::iterator found;

Basically, the compiler can't know that map< Configuration<TT>, Configuration<TT> >::iterator is a type rather than, e.g. a member variable, unless you tell it. Anytime you are using a named type that depends on the template parameter(s) you must use typename (except in a few exceptional cases such as in the initialization list of a constructor).