Problem with list<T>::iterator

[Gaijinco]

This code doesn´t compile:

template <typename T>
ostream& operator<<(ostream& out, const list<T>& l) {

list<T>::iterator it = l.begin();
while(it != (l.end() - 1)) {
out << *it << " ";
++it;
}
if(l.size() > 0) {
out << *it;
}

return out;
}

My compiler says: "expected ';' before it"

Why I'm doing wrong?

Thanks.

 

[rishabh]

This code doesn´t compile:

template <typename T>
ostream& operator<<(ostream& out, const list<T>& l) {

   list<T>::iterator it = l.begin();
   while(it != (l.end() - 1)) {
      out << *it << " ";
      ++it;
   }
   if(l.size() > 0) {
      out << *it;
   }

   return out;

}

My compiler says: "expected ';' before it"

Why I'm doing wrong?

Thanks.

Please include #include <list>

 

[peter koch]

This code doesn´t compile:

template <typename T>
ostream& operator<<(ostream& out, const list<T>& l) {

   list<T>::iterator it = l.begin();

You need a typename here: typename list<T>:......

Since this is templated code, you need to tell the compiler that
iterator is a type.

   while(it != (l.end() - 1)) {

And this will not compile - list iterators are not random.

      out << *it << " ";
      ++it;
   }
   if(l.size() > 0) {
      out << *it;
   }

   return out;

}

/Peter

 

[Peter Remmers]

And this will not compile - list iterators are not random.

And even if they were, the condition assumes that the list contains at
least one element, and fails if the list is empty (probably leads to the
condition never becoming true, at best).

/Peter

/Peter

 

[peter koch]

And even if they were, the condition assumes that the list contains at
least one element, and fails if the list is empty (probably leads to the
condition never becoming true, at best).

Yes - that is true as well - I focused on the syntax errors. The
entire design could be argued. While I might have a function like the
above for convenience, my central algorithm would be based on a pair
of iterators.

/Peter

/Peter