Puzzeled by Range object

ruby-talk
1

I was just writing an example for the “power of ruby” thread, using ranges.

I had two calls

b: (78…1).each {|y| print "#{y}, "}

The first worked as expected, it printed 1 to 78 to the screen.

However the second, simply printed nothing. Am I wrong to expect that a
range can work for both up and down?

(Using Ruby 1.6.6 windows installer, on Windows XP)

Rob

···

a: (1…78).each {|x| print "#{x}, "}

2

However the second, simply printed nothing. Am I wrong to expect that a
range can work for both up and down?

Range can work only up.

Except for Fixnum and Numeric, Range#each use #succ and it's difficult to
define the inverse of #succ for any type

Guy Decoux

3

I was just writing an example for the “power of ruby” thread, using ranges.

I had two calls

a: (1…78).each {|x| print "#{x}, "}
b: (78…1).each {|y| print "#{y}, "}

The first worked as expected, it printed 1 to 78 to the screen.

However the second, simply printed nothing. Am I wrong to expect that a
range can work for both up and down?

Yes.

(Using Ruby 1.6.6 windows installer, on Windows XP)

Rob

See:

http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/39324

http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/8993

-Martin

···

On Thu, Aug 22, 2002 at 12:12:30AM +0900, Robert McGovern wrote:

4

However the second, simply printed nothing. Am I wrong to expect that a
range can work for both up and down?

Ranges works in upward direction only … and while I don’t know the
reason, I think it’s the right behaviour. Intuitive explanation (for
me) is:

(1…5) <=> “{ x | 1 < x && x < 5 }”

thus:

(5…1) <=> “{ x | 5 < x && x < 1 }”

W.

···

–
Wejn <lists+rubytalk(at)box.cz>
(svamberk.net’s Linux section, fi.muni.cz student, linuxfan)

    Bored?  Want hours of entertainment?         <<<
      Just set the initdefault to 6!             <<<
5

What about this?

batsman@kodos:~/src/rubylang$ cat myrange.rb

class Range
def each
if (first <=> last) == -1
# proceed in normal order
a = first
while ( a != last )
yield a
a = a.succ
end
yield last unless exclude_end?
else
# reverse!
a = last
a = a.succ if exclude_end?
while ( a != first )
yield a
a = a.succ
end
end
end
end

l = [ “(1…3).each { |x| puts x }”,
“(-3…1).each { |x| puts x }”,
“(1…3).each { |x| puts x }”,
“(-3…1).each { |x| puts x }” ]

l.each { |x| puts “Executing #{x}”; eval x }
batsman@kodos:~/src/rubylang$ ruby myrange.rb
Executing (1…3).each { |x| puts x }
1
2
3
Executing (-3…1).each { |x| puts x }
-3
-2
-1
0
1
Executing (1…3).each { |x| puts x }
1
2
Executing (-3…1).each { |x| puts x }
-3
-2
-1
0
batsman@kodos:~/src/rubylang$
Script terminado (Wed Aug 21 21:27:04 2002
)

It will surely break something, anybody cares to test it?

···

On Thu, Aug 22, 2002 at 01:03:07AM +0900, ts wrote:

However the second, simply printed nothing. Am I wrong to expect that a
range can work for both up and down?

Range can work only up.

Except for Fixnum and Numeric, Range#each use #succ and it’s difficult to
define the inverse of #succ for any type

–
_ _

__ __ | | ___ _ __ ___ __ _ _ __
'_ \ / | __/ __| '_ _ \ / ` | ’ \
) | (| | |__ \ | | | | | (| | | | |
.__/ _,|_|/| || ||_,|| |_|
Running Debian GNU/Linux Sid (unstable)
batsman dot geo at yahoo dot com

Because I don’t need to worry about finances I can ignore Microsoft
and take over the (computing) world from the grassroots.
– Linus Torvalds

6

Thu, 22 Aug 2002 04:27:06 +0900, Mauricio Fernández batsman.geo@yahoo.com pisze:

What about this?

batsman@kodos:~/src/rubylang$ cat myrange.rb

It breaks the property that (1…n) has n elements even for n==0.

The direction can’t be inferred from the order of ends. It must be
specified explicitly.

···

–
__("< Marcin Kowalczyk
__/ qrczak@knm.org.pl
^^ http://qrnik.knm.org.pl/~qrczak/

7

“Mauricio Fernández” wrote

What about this?

batsman@kodos:~/src/rubylang$ cat myrange.rb

class Range
def each
if (first <=> last) == -1

proceed in normal order

a = first
while ( a != last )

Better use * while ( a < last ) * or you
will get an inifinite loop if the range is
built with floats like (1 … 5.7)

8

However you might write come code such as

(start…10).each {|x| … }

I would expect such code to loop from start to 10 for a values of start
<= 10. For values greater than 10 I would not expect the code to run.

For my code to start ‘running backwards’ as it were would be very
counter intuative.

9

Then I’ll give another name for the new iterator with the other
semantic. Plus now “reverse” ranges are processed in order,
ie 1…-3 => 1 0 -1 -2 -3 instead of -3 -2 -1 0 1 as before.

Note that when doing (A…B).each2,
A.succ must be defined if A < B
B.succ must be defined if B >= A
That’s why (1…2.3).each2 and (1.1…-3).each2 both work
but not (1.1…3).each2 nor (1…-3.1).each2.

A <=> B (or the reverse) must be defined in both cases.

batsman@kodos:~/src/rubylang$ cat myrange.rb

class Range
def each2
if (first <=> last) == -1
# proceed in normal order
a = first
while (a <=> last) == -1
yield a
a = a.succ
end
yield last unless exclude_end? if (a <=> last) == 0
else
# reverse!
v =
a = last
a = a.succ if exclude_end?
while (a <=> first) == -1
v << a
a = a.succ
end
v << a if (a <=> first) == 0
v.reverse!
v.each { |x| yield x }
end
end
end

l = [ “(1…3).each2 { |x| print x, ’ ’ }”,
“(1…-3).each2 { |x| print x, ’ ’ }”,
“(1…3).each2 { |x| print x, ’ ’ }”,
“(1…-3).each2 { |x| print x, ’ ’ }”,
“(1…3.1).each2 { |x| print x, ’ ’ }”,
“(1…3.1).each2 { |x| print x, ’ ’ }”,
“(1.1…-3).each2 { |x| print x, ’ ’ }”,
“(1.1…-3).each2 { |x| print x, ’ ’ }”,
]

l.each { |x| print "Executing #{x}: "; eval x; puts }

batsman@kodos:~/src/rubylang$ ruby myrange.rb
Executing (1…3).each2 { |x| print x, ’ ’ }: 1 2 3
Executing (1…-3).each2 { |x| print x, ’ ’ }: 1 0 -1 -2 -3
Executing (1…3).each2 { |x| print x, ’ ’ }: 1 2
Executing (1…-3).each2 { |x| print x, ’ ’ }: 1 0 -1 -2
Executing (1…3.1).each2 { |x| print x, ’ ’ }: 1 2 3
Executing (1…3.1).each2 { |x| print x, ’ ’ }: 1 2 3
Executing (1.1…-3).each2 { |x| print x, ’ ’ }: 1 0 -1 -2 -3
Executing (1.1…-3).each2 { |x| print x, ’ ’ }: 1 0 -1 -2
batsman@kodos:~/src/rubylang$

···

On Thu, Aug 22, 2002 at 05:08:59PM +0900, Peter Hickman wrote:

However you might write come code such as

(start…10).each {|x| … }

I would expect such code to loop from start to 10 for a values of start
<= 10. For values greater than 10 I would not expect the code to run.

For my code to start ‘running backwards’ as it were would be very
counter intuative.

–
_ _

__ __ | | ___ _ __ ___ __ _ _ __
'_ \ / | __/ __| '_ _ \ / ` | ’ \
) | (| | |__ \ | | | | | (| | | | |
.__/ _,|_|/| || ||_,|| |_|
Running Debian GNU/Linux Sid (unstable)
batsman dot geo at yahoo dot com

People disagree with me. I just ignore them.
– Linus Torvalds, regarding the use of C++ for the Linux kernel