Question

I wrote these classes that represent a graph:

class Node{

  public:
    typedef std::vector<Node>::iterator iterator;
    iterator begin() {return neigh.begin();}
    iterator end() {return neigh.end();}

  public:
    Node(std::string n) : name(n) {}

    std::string getName() {return name; }

    void addNeigh(Node n){
      neigh.push_back(n);
    } 

  private:
    std::string name;
    std::vector<Node> neigh;
};

class Map{

  public:
    typedef std::vector<Node>::iterator iterator;
    iterator begin() {return nodes.begin();}
    iterator end() {return nodes.end();}

  public:
    Map(std::string n) : map_name(n) {}

    void addNode(Node n){
      nodes.push_back(n);
    }


    std::string getName() { return map_name; }
    int getNumNodes() {return nodes.size();}

  private:
    std::string map_name;
    std::vector<Node> nodes;

};

Then i want to write a function printGraph that prints some graph informations using a traits:

template<typename T>
struct GraphTraits{

};

template<>
struct GraphTraits<Map>{

  typedef Map::iterator node_iterator;

  static node_iterator node_begin(Map map){ return map.begin();}
  static node_iterator node_end(Map map) { return map.end();}

  static std::string getName(Map m) {return m.getName();}

};

template<typename T>
void printGraph(T t){
  std::cout << std::endl 
            << "--------------------------------------" << std::endl  
            << "|             printGraph             |" << std::endl  
            << "--------------------------------------" << std::endl;
  std::cout << "Graph name: " << GraphTraits<T>::getName(t) << std::endl;

  for(GraphTraits<T>::node_iterator it = GraphTraits<T>::node_begin(t), e = GraphTraits<T>::node_end(t);
        it != e; ++it) {
          std::cout<< (*it).getName() << std::endl; }
}

If i try to compile this code i get these errors:

grafo.cpp:75:37: error: expected ';' in 'for' statement specifier
  for(GraphTraits<T>::node_iterator it = GraphTraits<T>::node_begin(t), e = GraphTraits<T>::node_end(t);
                                    ^
grafo.cpp:75:37: error: use of undeclared identifier 'it'
grafo.cpp:75:73: error: use of undeclared identifier 'e'
  for(GraphTraits<T>::node_iterator it = GraphTraits<T>::node_begin(t), e = GraphTraits<T>::node_end(t);
                                                                        ^
grafo.cpp:76:9: error: use of undeclared identifier 'it'
        it != e; ++it) {
        ^
grafo.cpp:76:15: error: use of undeclared identifier 'e'
        it != e; ++it) {
              ^
grafo.cpp:76:16: error: expected ')'
        it != e; ++it) {
               ^
grafo.cpp:75:6: note: to match this '('
  for(GraphTraits<T>::node_iterator it = GraphTraits<T>::node_begin(t), e = GraphTraits<T>::node_end(t);
     ^
grafo.cpp:76:20: error: use of undeclared identifier 'it'
        it != e; ++it) {
                   ^
grafo.cpp:75:23: error: unexpected type name 'node_iterator': expected expression
  for(GraphTraits<T>::node_iterator it = GraphTraits<T>::node_begin(t), e = GraphTraits<T>::node_end(t);
                      ^
grafo.cpp:104:3: note: in instantiation of function template specialization 'printGraph<Map>' requested here
  printGraph(mappa);
  ^

I can solve and compile successfully substituting:

for(GraphTraits<T>::node_iterator....

with:

for(GraphTraits<Map>::node_iterator....

But this makes printGraph to loose it’s generality.

Can someone give me some hint to solve the problem?

Talking in generale, is it possible to use a Traits to force a data-type definition? I mean:

template<typename T>
struct ATraits{
 -> type "iterator" must be defined
};

such that every specialization has to define:

typedef ..something.. iterator;

Answer 1

You forget to use the typename keyword.

In GraphTraits<Map>::node_iterator, the compiler knows what specialization of GraphTraits is used, and can figure out that node_iterator is a typedef.

In GraphTraits<T>::node_iterator, it has no idea what specialization will be used, because T isn’t known yet. So it assumes that node_iterator is a member variable, not a type. You have to correct this assumption. Say typename GraphTraits<T>::node_iterator

To require every specialization of GraphTraits to provide a typedef, you’d need “concepts”, which were proposed for C++11 but had to be delayed until a future version. So it isn’t possible in C++ today. Whenever you actually USE printGraph (formally, during instantiation of each template function instance of the printGraph<T> function template), however, the compiler will double-check that GraphTraits<T>::node_iterator really is a type (since now it knows T and can perform overload resolution).