C++ Using a class template argument as a template argument for another type

Question

Hey Everyone,

I'm having this problem while writing my own HashTable. It all works, but when I try to templatize the thing, it gave me errors. I recreated the problem as follows:

THIS CODE WORKS:

typedef double Item;

class A
{
public:
    A()
    {
        v.push_back(pair<string, Item>("hey", 5.0));
    }

    void iterate()
    {
        for(Iterator iter = v.begin(); iter != v.end(); ++iter)
            cout << iter->first << ", " << iter->second << endl;
    }

private:
    vector<pair<string, double> > v;
    typedef vector< pair<string, double> >::iterator Iterator;
};

THIS CODE DOES NOT:

template<typename ValueType>
class B
{
public:
    B(){}

    void iterate()
    {
        for(Iterator iter = v.begin(); iter != v.end(); ++iter)
            cout << iter->first << ", " << iter->second << endl;
    }

private:
    vector<pair<string, ValueType> > v;
    typedef vector< pair<string, ValueType> >::iterator Iterator;
};

the error messages: g++ -O0 -g3 -Wall -c -fmessage-length=0 -omain.o ..\main.cpp

..\main.cpp:50: error: type std::vector<std::pair<std::string, ValueType>, std::allocator<std::pair<std::string, ValueType> > >' is not derived from typeB'

..\main.cpp:50: error: ISO C++ forbids declaration of `iterator' with no type

..\main.cpp:50: error: expected `;' before "Iterator"

..\main.cpp: In member function `void B::iterate()':

..\main.cpp:44: error: `Iterator' was not declared in this scope

..\main.cpp:44: error: expected `;' before "iter"

..\main.cpp:44: error: `iter' was not declared in this scope

Does anybody know why this is happening? Thanks!

Answer 1

This is called "dependent names" in C++. In your second code snippet, you say:

typedef vector< pair<string, ValueType> >::iterator Iterator;

whereas you should say:

typedef typename vector< pair<string, ValueType> >::iterator Iterator;

Whenever you see an error saying "is not derived...", typename is to the rescue. In general, the idea is that the compiler doesn't know if iterator is a type or a variable, because it doesn't know what

vector< pair <string, ValueType> >

is, as it depends on ValueType.

(Afraid I'm not using the right terms here, but the idea is correct)

Answer 2

thanks it works and all makes sense now!

And i don't want to use a typedef, I can also use typename like this:

for( typename list<pair<string, ValueType> >::const_iterator itr = v.begin(); itr != v.end() ; ++itr){
{
//code
}