c++ template specialisation method definition - c++

Question

The following code works fine, a simple template class with a definition and a use
#include <string>
#include <iostream>
using namespace std;
template<class T> class foo{
public:
string what();
};
template<class T> string foo<T>::what(){
return "foo of type T";
}
int main(){
foo<int> f;
cout << f.what() << endl;
}
If I then add the following (above main, but after the declaration of template class foo;)
template<> class foo<char>{
public:
string what();
};
template<> string foo<char>::what(){
return "foo of type char";
}
I get an error from g++
Line 19: error: template-id 'what<>'
for 'std::string foo::what()'
does not match any template
declaration
Here is a codepad the shows the error: http://codepad.org/4HVBn9oJ
What obvious misstake am I making? Or is this not possible with c++ templates?
Will defining all the methods inline (with the defintion of template<> foo) work?
Thanks again all.

Answer 1

template<> class foo<char>{
public:
string what();
};
/*template<>*/ string foo<char>::what(){
return "foo of type char";
}
You don't need that template<>. foo<char> is already a complete type after it's specialized.

Answer 2

Writing this as:
#include <string>
#include <iostream>
using namespace std;
template<class T> class foo{
public:
string what();
};
template<class T> string foo<T>::what(){
return "foo of type T";
}
template<> class foo<char>{
public:
string what();
};
string foo<char>::what(){
return "foo of type char";
}
int main(){
foo<char> f;
cout << f.what() << endl;
}
works as expected.

Answer 3

If I then add the following (above main, but before the declaration of template class foo;)
Define the specialization after the generic class template.
By the time the compiler sees the specialization, it first needs to know the class template of which this is a specialization. So logically, the specialization should appear after the generic class template.