C++ confused with partial template specialization syntax - c++

Question

Is there a way I could avoid using Range2 as a name and have both classes named Range? I'm a bit confused with the C++ template syntax.
template <int BEGIN, int END>
class Range2
{
public:
class Iterator
{
private:
int n;
public:
Iterator(int n)
: n(n)
{
}
int operator *() const
{
return n;
}
Iterator const & operator ++()
{
++n;
return *this;
}
bool operator !=(Iterator const & i) const
{
return n != i.n;
}
};
Iterator begin() const
{
return Iterator(BEGIN);
}
Iterator end() const
{
return Iterator(END);
}
};
template<int END>
class Range
: public Range2<0, END>
{
};

Answer 1

As with function arguments, in C++ template arguments can have a default value. The only thing you will have to pay for this putting the END, which has no default value, before the BEGIN, which has 0 as default.
// Here we add the default parameter to BEGIN
// The arguments are switched because END is without a default
// parameter, so it has to come before BEGIN that has one
template <int END, int BEGIN=0>
// The class itself is the same, but now you can use it
// without giving a BEGIN parameters
class Range
{
public:
class Iterator
{
private:
int n;
public:
Iterator(int n)
: n(n)
{
}
int operator *() const
{
return n;
}
Iterator const & operator ++()
{
++n;
return *this;
}
bool operator !=(Iterator const & i) const
{
return n != i.n;
}
};
Iterator begin() const
{
return Iterator(BEGIN);
}
Iterator end() const
{
return Iterator(END);
}
};
It compiles and should work as intended. Without a main, I wasn't able to test it, though.
EDIT: I added some comments and here an example of usage, just for clarity:
Range<10, 3> r(3); /*here we use it as usual, pay attention begin is
now the second argument and not the first */
Range<10> r(0); /* here we don't give a BEGIN argument, the
compiler will use the default one */