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

[Jian Lin]

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.

 

[Rajinder Yadav]

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

 

[Jian Lin]

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

 

[Rob Biedenharn]

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

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 Biedenharn http://agileconsultingllc.com
(e-mail address removed)

 

[Jian Lin]

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

 

[botp]

for i =3D 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

=A0do something
next

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

class Fixnum
def countdown_by decrement =3D 1
x =3D self
x.times do |y|
yield(x - decrement*y)
end
end
end

5.countdown_by {|x| puts x}

5
4
3
2
1
=3D> 5

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

5
3
1
-1
-3
=3D> 5

kind regards -botp

 

[Rob Biedenharn]

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

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 Biedenharn http://agileconsultingllc.com
(e-mail address removed)

 

[Thairuby TH]

#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).

 

[Jian Lin]

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 =)