function template problem

[John Harrison]

This might be a compiler problem or it might be some subtlety of the
language I don't understand.

I'm writing an iterator and I want to write a version of std::distance
specially for my iterator. Here's some code (highly reduced of course). The
issue is how the return type of std::distance should be written.

#include <iostream>
#include <cstddef>
#include <iterator>

template <class T>
class Iter
{
public:
typedef std::input_iter ator_tag iterator_catego ry;
typedef T value_type;
typedef T& reference;
typedef T* pointer;
typedef std::size_t size_type;
typedef std::ptrdiff_t difference_type ;
bool operator==(Iter rhs) const { return true; }
bool operator!=(Iter rhs) const { return false; }
Iter& operator++() { return *this; }
Iter operator++(int) { return *this; }
};

namespace std
{
template <class T>
typename std::iterator_t raits<Iter<T> >::distance_typ e /* return type 1 */
// std::ptrdiff_t /* return type 2 */
distance(Iter<T > first, Iter<T> last)
{
std::cout << "success\n" ;
return 0;
}
}

int main()
{
std::distance(I ter<int>(), Iter<int>());
}

On MSVC++ 7.1 this prints "success" proving that my version of std::distance
has been called. On gcc 3.3.1 it does not. However if I change the return
type of my version std::distance to std::ptrdiff_t, then gcc 3.3.1 prints
"success" as well.

I was pretty much sure this was a gcc compiler bug but I thought I'd test
with Comeau C++ online, from what I've been able to tell Comeau works the
same as gcc, so now I'm not so sure.

Anyone able to tell me what's going on?

John

[Siemel Naran]

"John Harrison" <jo************ *@hotmail.com> wrote in message
news:2i******** ****@uni-berlin.de...

This might be a compiler problem or it might be some subtlety of the
language I don't understand.

I'm writing an iterator and I want to write a version of std::distance
specially for my iterator. Here's some code (highly reduced of course). The issue is how the return type of std::distance should be written.

#include <iostream>
#include <cstddef>
#include <iterator>

template <class T>
class Iter
{
public:
typedef std::input_iter ator_tag iterator_catego ry;
typedef T value_type;
typedef T& reference;
typedef T* pointer;
typedef std::size_t size_type;
typedef std::ptrdiff_t difference_type ;
bool operator==(Iter rhs) const { return true; }
bool operator!=(Iter rhs) const { return false; }
Iter& operator++() { return *this; }
Iter operator++(int) { return *this; }
};

namespace std
{
template <class T>
typename std::iterator_t raits<Iter<T> >::distance_typ e /* return type 1 */ // std::ptrdiff_t /* return type 2 */
distance(Iter<T > first, Iter<T> last)
{
std::cout << "success\n" ;
return 0;
}
}

int main()
{
std::distance(I ter<int>(), Iter<int>());
}

On MSVC++ 7.1 this prints "success" proving that my version of std::distance has been called. On gcc 3.3.1 it does not. However if I change the return
type of my version std::distance to std::ptrdiff_t, then gcc 3.3.1 prints
"success" as well.
MSVC 7 is right.

As an aside, I think you can define function distance in the same namespace
as class Iter. Koening lookup ensures that a call distance(iter1, iter2)
finds distance in the namespace where iter1 and iter2 live, and then in the
current namespace. But of course, people may write std::distance(. ..) which
forces the use of the function in the std namespace. So what you have still
seems a good idea.
I was pretty much sure this was a gcc compiler bug but I thought I'd test
with Comeau C++ online, from what I've been able to tell Comeau works the
same as gcc, so now I'm not so sure.

Anyone able to tell me what's going on?

There is a g++ newsgroup out there. Might be a bug.

[Rob Williscroft]

John Harrison wrote in news:2i******** ****@uni-berlin.de in
comp.lang.c++:

This might be a compiler problem or it might be some subtlety of the
language I don't understand.
typedef std::ptrdiff_t difference_type ;
template <class T>
typename std::iterator_t raits<Iter<T> >::distance_typ e /* return type
Anyone able to tell me what's going on?

SFINAE:

template<class T> struct iterator_traits <T*> {
typedef ptrdiff_t difference_type ;
typedef T value_type;
typedef T* pointer;
typedef T& reference;
typedef random_access_i terator_tag iterator_catego ry;
};

There is no distance_type, change to difference_type .
Rob.
--
http://www.victim-prime.dsl.pipex.com/

[John Harrison]

"Rob Williscroft" <rt*@freenet.co .uk> wrote in message
news:Xn******** *************** ***********@130 .133.1.4...

John Harrison wrote in news:2i******** ****@uni-berlin.de in
comp.lang.c++:

This might be a compiler problem or it might be some subtlety of the
language I don't understand.

typedef std::ptrdiff_t difference_type ;

template <class T>
typename std::iterator_t raits<Iter<T> >::distance_typ e /* return type

Anyone able to tell me what's going on?

SFINAE:

template<class T> struct iterator_traits <T*> {
typedef ptrdiff_t difference_type ;
typedef T value_type;
typedef T* pointer;
typedef T& reference;
typedef random_access_i terator_tag iterator_catego ry;
};

There is no distance_type, change to difference_type .

Aarrgh!! MSVC++ has distance_type for backwards compatibility.

Thanks, and well spotted, everything works now.

john