Leonid Leonid - 2 months ago 9
C++ Question

Overloading output stream operator for vector<T>

What is a recommended way to overload the output stream operator? The following can not be done. It is expected that compilation will fail if the operator << is not defined for a type T.

template < class T >
inline std::ostream& operator << (std::ostream& os, const std::vector<T>& v)
{
os << "[";
for (std::vector<T>::const_iterator ii = v.begin(); ii != v.end(); ++ii)
{
os << " " << *ii;
}
os << " ]";
return os;
}


EDIT: It does compile, the problem was unrelated and was in the namespace. Thanks for assistance.

Nim Nim
Answer

Did you actually try this code? It works fine on gcc with a small tweak std::vector<T>::const_iterator, needs to be declared as typename std::vector<T>::const_iterator

You may be better off with using std::copy and std::ostream_iterator.

EDIT: types, dependent types and typename Can't fit it all in the comments, so here goes (btw. this is my understanding and I could be off by a country mile - if so please correct me!)...

I think this is best explained with a simple example..

Let's assume you have a function foo

template <typename T>
void foo()
{
  T::bob * instofbob; // this is a dependent name (i.e. bob depends on T)
};

Looks okay, and typically you may do this

class SimpleClass
{
  typedef int bob;
};

And call

foo<SimpleClass>(); // now we know that foo::instofbob is "int"

Again, seems self explanatory, however some nuser comes along and does this

class IdiotClass
{
  static int bob;
};

Now

foo<IdiotClass>(); // oops, 

What you have now is an expression (multiplication) as IdiotClass::bob resolves to a non-type!

To the human, it's obvious that this is stupid, but the compiler has no way of differentiating between types vs. non-types, and by default in C++ (and I think this is where compilers differ), all qualified dependent names (i.e. T::bob) will be treated as non-type. To explicitly tell the compiler that the dependent name is a real type, you must specify the typename keyword -

template <typename T>
void foo()
{
  typedef typename T::bob *instofbob; // now compiler is happy, it knows to interpret "bob" as a type (and will complain otherwise!)
};

This applies even if it is a typedef. i.e.

template <typename T>
void foo()
{
  typedef typename T::bob local_bob;
};

Is that any clearer?