CODEWITHSUNDEEP
arraysruby
BrawdSword
BrawdSword

Reputation: 23

Ruby: Why is the last element In Array being changed to 1?

My function converts an input binary number into its decimal equivalent. However, when I enter 110 instead of getting 6 ([2^2 * 1, 2^1 * 1, 2^0 * 0] ⇒ [4 + 2 + 0] ⇒ 6) I get 7 because the last value in the Array is changed to 1. ([1, 1, 0] ⇒ [1, 1, 1]). This happens when I save the value at that index using .to_i to valueInt. At the same time, however, there is an if statement that should prevent anything being done to an element with a 0 in it. I am stumped.

puts "BinaryArr: #{binaryArr}"

# loops through the array in reverse order and converts the 1's
# to their respective weight by location
binaryArr.reverse.each_index do |index|
  if binaryArr[index] != 0 then
    valueInt = binaryArr[index].to_i
    puts "ValueInt: #{valueInt}"

    val = (2**index)*valueInt
    puts "val: #{val}"

    binaryArr[(binaryArr.length - 1) - index] = "#{val}"
  end
end

puts "WeightArr: #{binaryArr}"

# loops through the arr adding the elements together for the total (decimal value)
decimalInt = 0

binaryArr.each_index do |x|
  thisVal = binaryArr[x].to_i
  decimalInt += thisVal

  if binaryArr[x + 1] == nil then
    break
  end
end

puts "Decimal Value: #{decimalInt}"

THE OUTPUT: when 110 is input

Binary is: 3 Bits long

BinaryArr: ["1", "1", "0"]

ValueInt: 1    <== why is this a 1 instead of a 0!?!?

val: 1

ValueInt: 1

val: 2 

ValueInt: 1

val: 4

WeightArr: ["4", "2", "1"]

Decimal Value: 7

Upvotes: 2

Views: 64

Answers (2)

leobelizquierdo
leobelizquierdo

Reputation: 1668

You are indexing the binaryArr inside the loop binaryArr.reverse.each_index do... end to get the value valueInt, but you should be indexing the reverse. You can set the reverse in a variable like this reverseArr = binaryArr.reverse before the loop and make sure you iterate over reverseArr and indexing it inside the loop. For example:

puts "BinaryArr: #{binaryArr}"

reverseArr = binaryArr.reverse

# lops through the array in reverse order and converts the 1's
# to there respective weight by location
reverseArr.each_index do |index|
  valueInt = reverseArr[index].to_i
  puts "ValueInt: #{valueInt}" 
  val = (2**index)*valueInt
  puts "val: #{val}"
  binaryArr[(binaryArr.length-1)-index] = "#{val}"
end
puts "WeightArr: #{binaryArr}"
# loops through the arr adding the elements together for the total(decimal value)
decimalInt = 0
binaryArr.each_index do |x|
  thisVal = binaryArr[x].to_i
  decimalInt += thisVal
  if binaryArr[x+1]==nil then
    break
  end

end

puts "Decimal Value: #{decimalInt}"

Of course, you can remove this line: if binaryArr[index]!=0 then ... end

Upvotes: 2

steenslag
steenslag

Reputation: 80065

  • if binaryArr[index]!=0 then According to the output binaryArr contains strings. "a_string" != 0 will aways be true.
  • The result of 0**0 is 1.

Upvotes: 0

Related Questions