I created a class for prime numbers as so: class Primes def initialize end def prime?(number) # Method returns true if number is prime. limit = Math.sqrt(number).ceil flag = true if number % 2 == 0 flag = false else 3.step(limit,2) {|i| if number % i == 0 flag = false break end } end return flag end def show_primes(lower, upper) # Prints all primes between lower and upper # Lower is incremented 1 if it is even. # The arcane "(((lower/2).floor)*2+1)" performs this task (((lower/2).floor)*2+1).step(upper,2) {|i| if prime?(i) == true print i.to_s + " " end } end end Then when I enter a=primes.new a.show_primes(1000000,1000100) I get 1000003 1000033 1000037 1000039 1000081 1000099 1000001 Where is that trailing 1000001 coming from? It is not a prime number and in fact is the lower limit. If I enter the same methods outside of class Primes and enter show_primes(1000000,1000100) I don't get the lower limit at the end of the printed values. I am running version 1.18.4 using scite in Ubuntu.

on 2006-12-27 02:21

on 2006-12-27 03:53

Hi Charles, I entered the code in Ruby 1.8.5 in Windows, using Scintilla IDE :-)) (the joke is that Scintilla-Scite is not an IDE at all!) it gave the error: >ruby prime.rb prime.rb:34: undefined local variable or method `primes' for main:Object (NameError) Then changed a=primes.new a.show_primes(1000000,1000100) to a=Primes::new a.show_primes(1000000,1000100) The good result appeared :-) >ruby prime.rb 1000003 1000033 1000037 1000039 1000081 1000099 >Exit code: 0 Don't know how you had not an error instead of the wrong result before... For that you need a Ruby guru :-) Kind Regards J. Augusto

on 2006-12-27 04:09

Hi -- On Wed, 27 Dec 2006, Charles A Gray wrote: > Then when I enter > a=primes.new (As another respondent pointed out, you need Primes.new.) > a.show_primes(1000000,1000100) I get > 1000003 1000033 1000037 1000039 1000081 1000099 1000001 > Where is that trailing 1000001 coming from? It is not a prime number and > in fact is the lower limit. > If I enter the same methods outside of class Primes and enter > show_primes(1000000,1000100) I don't get the lower limit at the end of > the printed values. > > I am running version 1.18.4 using scite in Ubuntu. I can't duplicate the problem, I'm afraid. David

on 2006-12-27 04:25

On Dec 26, 2006, at 8:20 PM, Charles A Gray wrote: > if number % 2 == 0 > end > end > the printed values. > > I am running version 1.18.4 using scite in Ubuntu. > -- > Charles Gray -- Phoenix, AZ; Where you can bake the chill out of your > bones The value of the #step method is the initial value so that is being returned from a.show_primes and presumably printed by whatever you're using to interpret your statements (like irb): With your code in a file names "primes.rb" $ irb -rprimes >> a=Primes.new => #<Primes:0x6e5d88> >> a.show_primes(1_000_000, 1_000_100) 1000003 1000033 1000037 1000039 1000081 1000099 => 1000001 >> a.show_primes(1_000_000, 1_000_100); nil 1000003 1000033 1000037 1000039 1000081 1000099 => nil >> a.show_primes(1_000_000, 1_000_100); puts "" 1000003 1000033 1000037 1000039 1000081 1000099 => nil Note that the value of the last expression is displayed by irb itself. (In the last example, the puts supplies a newline and the value of puts as an expression is nil.) To see the documentation for the step method (after first trying Fixnum#step and Integer#step rather than looking it up in the pickaxe ;-) $ ri -T Numeric#step ----------------------------------------------------------- Numeric#step num.step(limit, step ) {|i| block } => num ------------------------------------------------------------------------ Invokes _block_ with the sequence of numbers starting at _num_, incremented by _step_ on each call. The loop finishes when the value to be passed to the block is greater than _limit_ (if _step_ is positive) or less than _limit_ (if _step_ is negative). If all the arguments are integers, the loop operates using an integer counter. If any of the arguments are floating point numbers, all are converted to floats, and the loop is executed _floor(n + n*epsilon)+ 1_ times, where _n = (limit - num)/step_. Otherwise, the loop starts at _num_, uses either the +<+ or +>+ operator to compare the counter against _limit_, and increments itself using the +++ operator. 1.step(10, 2) { |i| print i, " " } Math::E.step(Math::PI, 0.2) { |f| print f, " " } _produces:_ 1 3 5 7 9 2.71828182845905 2.91828182845905 3.11828182845905 -Rob Rob Biedenharn http://agileconsultingllc.com Rob@AgileConsultingLLC.com

on 2006-12-27 04:43

On Dec 26, 2006, at 10:07 PM, dblack@wobblini.net wrote: >> 1000003 1000033 1000037 1000039 1000081 1000099 1000001 > I can't duplicate the problem, I'm afraid. > > > David > > -- > Q. What is THE Ruby book for Rails developers? > A. RUBY FOR RAILS by David A. Black (http://www.manning.com/black) > (See what readers are saying! http://www.rubypal.com/r4rrevs.pdf) > Q. Where can I get Ruby/Rails on-site training, consulting, coaching? > A. Ruby Power and Light, LLC (http://www.rubypal.com) Charles, I was prepared to be disheartened when I saw that David had responded ahead of me, but I have a solution! $ irb -f --noprompt -rprimes a=Primes.new #<Primes:0x1d5084> a.show_primes(1_000_000, 1_000_100) 1000003 1000033 1000037 1000039 1000081 1000099 1000001 exit Since your commands didn't show the typical prompts from irb, I went poking around and came up with this. The '-f' suppresses the ~/.irbrc (which I have), --noprompt ought to be obvious, and -rprimes requires the primes.rb file with your original Primes class definition. -Rob Rob Biedenharn http://agileconsultingllc.com Rob@AgileConsultingLLC.com

on 2006-12-27 04:45

On Wed, 2006-12-27 at 12:07 +0900, dblack@wobblini.net wrote: > > > Then when I enter > > a=primes.new > > (As another respondent pointed out, you need Primes.new.) That was a typo. I actually had a=Primes.new. Which Ruby are you running?

on 2006-12-27 05:11

On Wed, 2006-12-27 at 12:23 +0900, Rob Biedenharn wrote: > > limit = Math.sqrt(number).ceil > > end > > end > > If I enter the same methods outside of class Primes and enter > The value of the #step method is the initial value so that is being > 1000003 1000033 1000037 1000039 1000081 1000099 => 1000001 > To see the documentation for the step method (after first trying > incremented by _step_ on each call. The loop finishes when the > the +++ operator. > -Rob Rob, I went back and added the puts for a blank line and got the same results as before, namely: a=Primes.new p a.prime?(1000001) p a.show_primes(1000000,1000100);puts "" produced: >ruby primes.rb false 1000003 1000033 1000037 1000039 1000081 1000099 1000001 >Exit code: 0 I am still running it in scite. As well as being a Ruby newby, I am a Linux newby and haven't figured out how to run irb in Linux. I am going to reboot into windows and see what happens there.

on 2006-12-27 05:15

Rob Biedenharn wrote: > The value of the #step method is the initial value so that is being > returned from a.show_primes and presumably printed by whatever you're > using to interpret your statements (like irb): As a simple example: irb(main):001:0> puts 1.step( 2 ){ print "a " } a a 1 => nil In the above, the block is called twice, and prints out 'a ' each time. Then the return value of the step method (the initial value) it passed to the puts function, which prints it out. You're seeing your first non-even value as the return, since the step method is the last call in your show_primes method. One other comment - the pairing of methods you have inside your class seems odd, given Ruby's built-in classes for numbers. Instead of your class wrapper, I might personally place those methods as: class Integer def is_prime? # ... end end def show_primes( lower, upper ) end Although, since I'm personally not a big fan of 'global' functions, I might even do: class Integer def self.show_primes( lower, upper ) # ... end end

on 2006-12-27 05:27

On Wed, 2006-12-27 at 11:51 +0900, Jose Augusto wrote: > For that you need a Ruby guru :-) > > Kind Regards > > J. Augusto > Tahnks for the reply a=Primes.new ans a=Primes::new both give me the same incorret result

on 2006-12-27 05:31

On Dec 26, 2006, at 11:09 PM, Charles A Gray wrote: > I am still running it in scite. As well as being a Ruby newby, I am a > Linux newby and haven't figured out how to run irb in Linux. Opening a terminal and typing irb should just work assuming ruby is installed in a standard fashion. Where to find the terminal will vary a bit by distro. Worst case, you should be able to hit Ctrl+Alt +F1 and jump to a text terminal on any distro (should be Ctrl+Alt+F7 to get you back to X windows, possibly F11). Have fun! -Mat

on 2006-12-27 06:18

Charles A Gray wrote: > On Wed, 2006-12-27 at 11:51 +0900, Jose Augusto wrote: >> For that you need a Ruby guru :-) >> >> Kind Regards >> >> J. Augusto >> > Tahnks for the reply a=Primes.new ans a=Primes::new both give me the > same incorret result Try this: class Primes def prime?( number ) ( ( ( number / 2 ).floor ) * 2 + 1 ) end end a = Primes.new a.prime?( 1000000 ) The last variable assigned is being returned from a.prime? (in this case an anonymous variable). This is how Ruby returns a value from a method. Hope this helps. Joe