if there is a loop
35.times do |i|
print i
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.
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
Do Good ~ Share Freely
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 }
I don’t want to create an extra variable. It is a simple loop to
count down 35 times in a short program.
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, " "
end
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. 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
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
#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).
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 =)
On Tue, Oct 20, 2009 at 1:14 PM, Jian L. [email protected]
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
end5.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
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