converting a string to bits in ruby, adding 1 bit and converting back to ascii

Question

I am writing a ruby uid generator and i need to generate unique ids in incremental order. The unique ids are strings that vary from 3 to 9 characters.

What i want to achieve is this:

Get the last generated identifier, convert it to bits, and 1 bit to it(basically do a + 1) and convert the result back to ascii. This way i can prevent a unique id from being generated twice.

How can i do this in ruby. I am aware of #pack and #unpack methods but i can't figure out a way.

Could you give an example?

Stefan
– Stefan

2014-07-25 05:14:30 +00:00
Commented Jul 25, 2014 at 5:14 — Stefan
– Stefan, Commented Jul 25, 2014 at 5:14

mvidaurre · Accepted Answer · 2014-07-25 06:49:15Z

3

You can use something like this:

a = "Foobar".unpack("B*")[0].to_i(2)+1

=> 77444424032627

["0" + a.to_s(2)].pack("B*")

=> "Foobas"

answered Jul 25, 2014 at 6:49

mvidaurre

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4 Comments

Stefan Over a year ago

that's equivalent to "Foobar".succ

Leon Over a year ago

So what I need is this: I have a set of characters [0-9A-Z], su numbers and only uppercase letters. I want to be able to start from 000 and go up to ZZZZZZZZZ(9 Zs) just by changing the last bit without choosing randomly. Is that possible in ruby?

mvidaurre Over a year ago

Do you want all the permutations? Use then: a = (0..9).to_a + ("A".."Z").to_a a.permutation(3).to_a

mvidaurre Over a year ago

Or if you want a function that based on a uid then generates the successor uid, then you should use succ and some logic in your function for detecting changes for numbers and chars. Explore this code: a = "0A"; (1..(10**2)).each do |m|; puts a; a.succ!; end

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