Can 35.times do |i| be decrement of "i"?

ruby-talk
1

if there is a loop

35.times do |i|
  print i
  # do something here...
end

it is nice except i hope to count down the "i"...

it can be

  print 35 - i

but then if the number needs to be 36 then i need to change both places.

or it can be

35.downto(1) do |i|

but doing so feel like needing to do some counting... 1 to 35 is in fact
35 times... so that's good

if 35.times can go decrement that would be nice too.

thanks.

···

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2

Jian Lin wrote:

if there is a loop

35.times do |i|
  print i
  # do something here...
end

it is nice except i hope to count down the "i"...

it can be

  print 35 - i

but then if the number needs to be 36 then i need to change both places.

or it can be

35.downto(1) do |i|

but doing so feel like needing to do some counting... 1 to 35 is in fact
35 times... so that's good

if 35.times can go decrement that would be nice too.

thanks.

Not sure I follow what you're asking, or why downto is not a good solution? but you could try this:

irb(main):008:0> count=5
irb(main):009:0> count.times { |n| puts count-n }
5
4
3
2
1
=> 5

···

--
Kind Regards,
Rajinder Yadav

http://DevMentor.org
Do Good ~ Share Freely

3

#Ruby has "reverse_each" method.

irb(main):001:0> RUBY_VERSION
=> "1.9.1"

irb(main):002:0> 35.times {|i| print i, " "}
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
27 28 29
30 31 32 33 34 => 35

irb(main):003:0> 35.times.reverse_each {|i| print i, " "}
34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11
10 9 8 7
6 5 4 3 2 1 0 => #<Enumerator:0x142c840>

#Note that it counts 34 down to 0 (not 35 down to 1).

···

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4

Rajinder Yadav wrote:

Jian Lin wrote:

if 35.times can go decrement that would be nice too.

thanks.

Not sure I follow what you're asking, or why downto is not a good
solution? but
you could try this:

irb(main):008:0> count=5
irb(main):009:0> count.times { |n| puts count-n }

1. I don't want to create an extra variable. It is a simple loop to
count down 35 times in a short program.

2. I want 35.times because it says it is 35 times, very clearly.
35.downto(1) you will need to think a little how many times it is. My
purpose is to do it 35 times, so 35.times is best, but I need the count
down. Something like

35.times.countdown do |i|
  print i, " "
  # do something
end

···

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5

Thairuby TH wrote:

irb(main):003:0> 35.times.reverse_each {|i| print i, " "}
34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11
10 9 8 7
6 5 4 3 2 1 0 => #<Enumerator:0x142c840>

#Note that it counts 34 down to 0 (not 35 down to 1).

counting from 34 to 0 is better since it is good for rocket take off =)

···

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6

Rajinder Yadav wrote:

Jian Lin wrote:

if 35.times can go decrement that would be nice too.

thanks.

Not sure I follow what you're asking, or why downto is not a good
solution? but
you could try this:

irb(main):008:0> count=5
irb(main):009:0> count.times { |n| puts count-n }

1. I don't want to create an extra variable. It is a simple loop to
count down 35 times in a short program.

2. I want 35.times because it says it is 35 times, very clearly.
35.downto(1) you will need to think a little how many times it is. My
purpose is to do it 35 times, so 35.times is best, but I need the count
down. Something like

35.times.countdown do |i|
print i, " "
# do something
end

--
Posted via http://www.ruby-forum.com/\.

So DO IT that way. Ruby lets you!

class Integer

        def countdown
          self.downto(1){|i|yield i}
        end
      end
=> nil

35.countdown {|i| print i, ' '}; puts "Boom!"

35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 Boom!
=> nil

Happy?

-Rob

Rob Biedenharn http://agileconsultingllc.com
Rob@AgileConsultingLLC.com

···

On Oct 20, 2009, at 12:44 AM, Jian Lin wrote:

7

Rob Biedenharn wrote:

> class Integer
        def countdown
          self.downto(1){|i|yield i}
        end
      end
=> nil
> 35.countdown {|i| print i, ' '}; puts "Boom!"
35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13
12 11 10 9 8 7 6 5 4 3 2 1 Boom!

yup, that's similar to what i was looking for. And 35.times doesn't
have any mechanism to count down i guess, not like

for i = 35 to 1 step -1
  do something
next

···

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8

for i = 35 to 1 step -1

hmm if you modify the step var, the block would not loop 35 times, wc
you ask in your op

do something
next

yours was a special case, so i was thinking something like,

class Fixnum
  def countdown_by decrement = 1
    x = self
    x.times do |y|
      yield(x - decrement*y)
    end
  end
end

5.countdown_by {|x| puts x}

5
4
3
2
1
=> 5

5.countdown_by(2) {|x| puts x}

5
3
1
-1
-3
=> 5

kind regards -botp

···

On Tue, Oct 20, 2009 at 1:14 PM, Jian Lin <blueskybreeze@gmail.com> wrote:

9

Rob Biedenharn wrote:

> class Integer
       def countdown
         self.downto(1){|i|yield i}
       end
     end
=> nil
> 35.countdown {|i| print i, ' '}; puts "Boom!"
35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13
12 11 10 9 8 7 6 5 4 3 2 1 Boom!

yup, that's similar to what i was looking for. And 35.times doesn't
have any mechanism to count down i guess, not like

for i = 35 to 1 step -1
do something
next

-- Posted via http://www.ruby-forum.com/\.

Just use Integer#step if that's how you want to think about it:

35.step(1,-1) {|i| print i,' '}; puts "Ha!"

35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 Ha!
=> nil

-Rob

Rob Biedenharn http://agileconsultingllc.com
Rob@AgileConsultingLLC.com

···

On Oct 20, 2009, at 1:14 AM, Jian Lin wrote: