Decomposing an integer into an array of integers by bit manipulation

Question

I'd like to write a method that takes an integer and a bit length and returns an array of integers corresponding to the bits.

For example:

decompose(100, 4)
#=> [4, 12]

because:

100 is 01001100 in binary
        /   \
      0100  1100
       4      12

decompose(123456, 6)
#=> [1, 8, 60, 0]

because:

123456 is 000001001000111100000000 in binary
           /       |     |      \
          1        8     60      0

Note: I don't need to worry about bit lengths that aren't exact divisors.

I don't think you have the right number.

sawa
– sawa

2012-09-29 16:44:42 +00:00
Commented Sep 29, 2012 at 16:44 — sawa
– sawa, Commented Sep 29, 2012 at 16:44

sawa · Accepted Answer · 2012-09-29 16:52:11Z

2

def decompose n, l, a = []
  n, r = n.divmod(2 ** l)
  a.unshift(r)
  n.zero? ? a : decompose(n, l, a)
end

decompose(100, 4) # => [6, 4]
decompose(123456, 6) # => [30, 9, 0]

answered Sep 29, 2012 at 16:52

sawa

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Comments

Michael Kohl · Accepted Answer · 2012-09-29 16:46:10Z

0

How about this?

def decompose(num, len)
  num.to_s(2).chars.each_slice(len).map { |x| x.join.to_i(2) }.reverse
end

decompose(100, 4)
#=> [4, 12]

Or this:

def decompose(num, len)
  num.to_s(2).scan(/.{1,len}/).map { |x| x.to_i(2) }.reverse
end

answered Sep 29, 2012 at 16:46

Michael Kohl

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