whats wrong with the following code?

Code:

---

#include <stdlib.h>
#include <iostream>
#include <fstream>
#include <list>
using namespace std;

ostream& M::print(ostream& stream) {
        s::iterator i;
        for(i = t->begin(); i != t->end(); ++i)
        {
                stream << *i;
        }

}

---

?

Where is pointer t defined and what is the error message?

t is defined in the header file

Code:

---

.
.
.
private:
  typedef list<ML> s;
        s * t;

---

and i get a huge list of errors

Code:

---

M.cpp: In member function `std::ostream&
  M::print(std::ostream&)':
M.cpp:56: error: no match for 'operator<<' in 'stream <<
  (&i)->std::_List_iterator<_Tp, _Ref, _Ptr>::operator*() const [with _Tp =
  ML, _Ref = ML&, _Ptr = ML*]()'
/usr/include/c++/3.3.2/bits/ostream.tcc:63: error: candidates are:
  std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT,
  _Traits>::operator<<(std::basic_ostream<_CharT,
  _Traits>&(*)(std::basic_ostream<_CharT, _Traits>&)) [with _CharT = char,
  _Traits = std::char_traits<char>]
/usr/include/c++/3.3.2/bits/ostream.tcc:85: error:
  std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT,
  _Traits>::operator<<(std::basic_ios<_CharT,
  _Traits>&(*)(std::basic_ios<_CharT, _Traits>&)) [with _CharT = char, _Traits
  = std::char_traits<char>]
/usr/include/c++/3.3.2/bits/ostream.tcc:107: error:
  std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT,
  _Traits>::operator<<(std::ios_base&(*)(std::ios_base&)) [with _CharT = char,
  _Traits = std::char_traits<char>]
/usr/include/c++/3.3.2/bits/ostream.tcc:179: error:
  std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT,
  _Traits>::operator<<(long int) [with _CharT = char, _Traits =
  std::char_traits<char>]
/usr/include/c++/3.3.2/bits/ostream.tcc:216: error:
  std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT,
  _Traits>::operator<<(long unsigned int) [with _CharT = char, _Traits =
  std::char_traits<char>]
.
.
.
list goes on..

---

I´m not sure but shouldn't M::print() also be templated??

not sure..
but i've realised that when i have

Code:

---


ostream& M::print(ostream& stream) {
        s::iterator i;
        for(i = t->begin(); i != t->end(); ++i)
        {
                stream << *i;
        }
}

---

that gives me a bunch of errors
but if i have

Code:

---

ostream& M::print(ostream& stream) {
        s::iterator i;
        for(i = t->begin(); i != t->end(); ++i)
        {
                stream << "hello";
        }
}

---

that gives me no problems
so im thinking that it could be the *i...
i've just been searching aroudn and i foudn this

Code:

---

if (! i->passed())    // iterators also provide operator ->
      {
        cout << "The student " << *i << " failed." << endl;
          // provided that class Student provides the overloaded
          // stream insertion operator <<
      }

---

not sure what that means tho..

just out of curosity what happen when you instantiate your class (M) template with a primitive datatype?

what do you mean? i dont understand..
im pretty new to c++

Hmm, pretty new to c++...I don´t want to be rude but templates (particular class) shoudln´t be learned to a "im pretty new to c++" in c++. You can´t run if you can´t walk :(. But I will make a try.

Primitive datatypes: Are datatypes that are built-in in c++ e.i int, float, double, char...

Template instantiation: Is when a function or a class is substituted with a buildt-in- or an user defined type.

hmmm yeah you've lost me.. haha
but now i'm defintely sure that the probelm is
when i do the printout of *i

i is the iterator for your list, which is of type ML. The ostream class doesn't know how to handle the '<<' operator with that class unless you define it. You'll need to overload the '<<' operator to work with an ostream and your ML class, or go about it by calling a member function of the ML class that will just print it's data types or something.

The FAQ and tutorials section will tell you how to go about overloading the '<<' operator (or the insertion operator in this case, if you're curious about a formal name...not to be confused with the bitshif left operator).