Re: Question on usage of Stream

>>>>> "Sebastien" == Sebastien Bocq writes:

Sebastien> Hello, The following simple example illustrates the usage of
Sebastien> streams in the documentation of Stream.scala:
>> def from(n: Int): Stream[Int] = Stream.cons(n, from(n + 1))
Sebastien> It looks very simple but the problem is that obviously the
Sebastien> method is not tail recursive and if I do:
>> from(2) take 3715 print
Sebastien> I get a java.lang.StackOverflowError
Sebastien> This shows that building infinite streams this way is quite
Sebastien> limited :)

A side note: with 2.7.2.final, I don't get a stack overflow error until
the 130154th item. I don't know your stack is overflowing so much
earlier.

More importantly: this works fine, no overflow:

scala> from(2).drop(100000000).first

which suggests to me that the source of the overflow in your example
is different than you think. It looks to me like the problem is in
Stream.print(), not in from() itself.

Indeed, James said so too. What he suggested is something like this:

scala> from(2) take 200000 foreach print

which works fine on my machine, so yes, its probably a problem with Stream.print. It is defined like this:

def print(sep: String) {
    if (isEmpty) Console.println("Stream.empty")
    else { Console.print(head); Console.print(sep); tail.print(sep) }
  }

Shouldn't there be a 'final' in front of the method?

Sebastien

2009/1/8 Seth Tisue <seth@tisue.net>

>>>>> "Sebastien" == Sebastien Bocq <sebastien.bocq@gmail.com> writes:

 Sebastien> Hello, The following simple example illustrates the usage of
 Sebastien> streams in the documentation of Stream.scala:
 >> def from(n: Int): Stream[Int] = Stream.cons(n, from(n + 1))
 Sebastien> It looks very simple but the problem is that obviously the
 Sebastien> method is not tail recursive and if I do:
 >> from(2) take 3715 print
 Sebastien> I get a java.lang.StackOverflowError
 Sebastien> This shows that building infinite streams this way is quite
 Sebastien> limited :)

A side note: with 2.7.2.final, I don't get a stack overflow error until
the 130154th item.  I don't know your stack is overflowing so much
earlier.

More importantly: this works fine, no overflow:

scala> from(2).drop(100000000).first

which suggests to me that the source of the overflow in your example
is different than you think.  It looks to me like the problem is in
Stream.print(), not in from() itself.

--
Seth Tisue / http://tisue.net
lead developer, NetLogo: http://ccl.northwestern.edu/netlogo/

>>>>> "Sebastien" == Sebastien Bocq writes:

Sebastien> Indeed, James said so too. What is suggest is something like
Sebastien> this:
scala> from(2) take 200000 foreach print

Sebastien> which works fine on my machine, so yes, its probably a
Sebastien> problem with Stream.print. It is defined like this:

Sebastien> def print(sep: String) { if (isEmpty)
Sebastien> Console.println("Stream.empty") else { Console.print(head);
Sebastien> Console.print(sep); tail.print(sep) } }

Sebastien> Shouldn't there be a 'final' in front of the method?

I imagine that would fix the problem, but it would also tie the hands of
people wanting to subclass Stream. I think the standard solution in
these cases is to nest a locally defined truly tail recursive function
inside the outer definition.

That makes more sense indeed, shall I file a bug?

Sebastien

2009/1/8 Seth Tisue <seth@tisue.net>

>>>>> "Sebastien" == Sebastien Bocq <sebastien.bocq@gmail.com> writes:

 Sebastien> Indeed, James said so too. What is suggest is something like
 Sebastien> this:
 scala> from(2) take 200000 foreach print

 Sebastien> which works fine on my machine, so yes, its probably a
 Sebastien> problem with Stream.print. It is defined like this:

 Sebastien> def print(sep: String) { if (isEmpty)
 Sebastien> Console.println("Stream.empty") else { Console.print(head);
 Sebastien> Console.print(sep); tail.print(sep) } }

 Sebastien> Shouldn't there be a 'final' in front of the method?

I imagine that would fix the problem, but it would also tie the hands of
people wanting to subclass Stream.  I think the standard solution in
these cases is to nest a locally defined truly tail recursive function
inside the outer definition.

--
Seth Tisue / http://tisue.net
lead developer, NetLogo: http://ccl.northwestern.edu/netlogo/

I created a ticket: http://lampsvn.epfl.ch/trac/scala/ticket/1631

Thanks,
Sebastien

2009/1/9 Sebastien Bocq <sebastien.bocq@gmail.com>

That makes more sense indeed, shall I file a bug?

Sebastien

2009/1/8 Seth Tisue <seth@tisue.net>

>>>>> "Sebastien" == Sebastien Bocq <sebastien.bocq@gmail.com> writes:

 Sebastien> Indeed, James said so too. What is suggest is something like
 Sebastien> this:
 scala> from(2) take 200000 foreach print

 Sebastien> which works fine on my machine, so yes, its probably a
 Sebastien> problem with Stream.print. It is defined like this:

 Sebastien> def print(sep: String) { if (isEmpty)
 Sebastien> Console.println("Stream.empty") else { Console.print(head);
 Sebastien> Console.print(sep); tail.print(sep) } }

 Sebastien> Shouldn't there be a 'final' in front of the method?

I imagine that would fix the problem, but it would also tie the hands of
people wanting to subclass Stream.  I think the standard solution in
these cases is to nest a locally defined truly tail recursive function
inside the outer definition.

--
Seth Tisue / http://tisue.net
lead developer, NetLogo: http://ccl.northwestern.edu/netlogo/