# Can 35.times do |i| be decrement of "i"? **URL:** \ **Category:** Ruby\ **Created:** [October 20, 2009, 6:10am UTC](https://www.ruby-forum.com/t/can-35-times-do-i-be-decrement-of-i/177556 "2009-10-20T06:10:42Z")\ **Posts on this page:** 9\ **Page:** 1
**Author:** ![winterheat](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/7993a0/32.png) [@winterheat](https://www.ruby-forum.com/u/winterheat)\ **Post date:** [October 20, 2009, 6:10am UTC](https://www.ruby-forum.com/t/can-35-times-do-i-be-decrement-of-i/177556/1 "2009-10-20T06:10:42Z")
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. ---
**Author:** ![winterheat](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/7993a0/32.png) [@winterheat](https://www.ruby-forum.com/u/winterheat)\ **Post date:** [October 20, 2009, 6:29am UTC](https://www.ruby-forum.com/t/can-35-times-do-i-be-decrement-of-i/177556/2 "2009-10-20T06:29:50Z")
Jian L. 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 } 5 4 3 2 1 =\> 5 – Kind Regards, Rajinder Y. [http://DevMentor.org](http://DevMentor.org) Do Good ~ Share Freely ---
**Author:** ![winterheat](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/7993a0/32.png) [@winterheat](https://www.ruby-forum.com/u/winterheat)\ **Post date:** [October 20, 2009, 6:44am UTC](https://www.ruby-forum.com/t/can-35-times-do-i-be-decrement-of-i/177556/3 "2009-10-20T06:44:32Z")
Rajinder Y. wrote: > Jian L. 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 ---
**Author:** ![winterheat](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/7993a0/32.png) [@winterheat](https://www.ruby-forum.com/u/winterheat)\ **Post date:** [October 20, 2009, 7:03am UTC](https://www.ruby-forum.com/t/can-35-times-do-i-be-decrement-of-i/177556/4 "2009-10-20T07:03:47Z")
On Oct 20, 2009, at 12:44 AM, Jian L. wrote: > > you could try this: > > purpose is to do it 35 times, so 35.times is best, but I need the So DO IT that way. Ruby lets you! irb\> class Integer def countdown self.downto(1){|i|yield i} end end =\> nil irb\> 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 B. [http://agileconsultingllc.com](http://agileconsultingllc.com) removed\_email\_address@domain.invalid ---
**Author:** ![winterheat](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/7993a0/32.png) [@winterheat](https://www.ruby-forum.com/u/winterheat)\ **Post date:** [October 20, 2009, 7:14am UTC](https://www.ruby-forum.com/t/can-35-times-do-i-be-decrement-of-i/177556/5 "2009-10-20T07:14:06Z")
Rob B. wrote: > irb\> class Integer > def countdown > self.downto(1){|i|yield i} > end > end > =\> nil > irb\> 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 ---
**Author:** ![winterheat](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/7993a0/32.png) [@winterheat](https://www.ruby-forum.com/u/winterheat)\ **Post date:** [October 20, 2009, 8:06am UTC](https://www.ruby-forum.com/t/can-35-times-do-i-be-decrement-of-i/177556/6 "2009-10-20T08:06:26Z")
On Oct 20, 2009, at 1:14 AM, Jian L. wrote: > > 12 11 10 9 8 7 6 5 4 3 2 1 Boom! Just use Integer#step if that’s how you want to think about it: irb\> 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 B. [http://agileconsultingllc.com](http://agileconsultingllc.com) removed\_email\_address@domain.invalid ---
**Author:** ![winterheat](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/7993a0/32.png) [@winterheat](https://www.ruby-forum.com/u/winterheat)\ **Post date:** [October 20, 2009, 9:03pm UTC](https://www.ruby-forum.com/t/can-35-times-do-i-be-decrement-of-i/177556/7 "2009-10-20T21:03:06Z")
#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). ---
**Author:** ![winterheat](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/7993a0/32.png) [@winterheat](https://www.ruby-forum.com/u/winterheat)\ **Post date:** [October 20, 2009, 10:50pm UTC](https://www.ruby-forum.com/t/can-35-times-do-i-be-decrement-of-i/177556/8 "2009-10-20T22:50:32Z")
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 =) ---
**Author:** ![winterheat](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/w/7993a0/32.png) [@winterheat](https://www.ruby-forum.com/u/winterheat)\ **Post date:** [October 20, 2009, 8:02am UTC](https://www.ruby-forum.com/t/can-35-times-do-i-be-decrement-of-i/177556/9 "2009-10-20T08:02:20Z")
On Tue, Oct 20, 2009 at 1:14 PM, Jian L. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote: > 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