float conversion logic

[Insectoid]

So, I just want some confirmation on my float conversion logic.

I am writing a program to convert a float to another number system. I have managed to separate the float into 2 parts (5.9 => 5, 9). My method for converting the decimal is to flip it over (053 => 350), then convert it in the same manner as an integer, using div and mod, but attach the result backwards again. Is this logic correct? so 50.05 converted to hex would be 32.23?

code, with comments at key points:

Ruby:

require 'jcode' class NumberConverter         attr_reader :decimal         def initialize (number, base)                 @number = number                 @base = base                 @integer = @number.to_i #the integer portion of the float                 @decimal = (@number.to_s.split(".")[1]).reverse.to_i #the decimal portion of the float, flipped         end                 def to_base(base)                 chars =("0".."9").to_a+("A".."Z").to_a                 answer = ""                 temp = @integer                 until (temp == 0) do                         answer = chars[(temp %base)] + answer #adds to front of answer                         temp /= base                 end                 temp = @decimal                 answer += "."                 until (temp == 0) do                         answer += chars[(temp %base)] #adds to back of answer                         temp /= base                 end                 answer.empty? ?  "0" : answer         end end puts (NumberConverter.new(50.05 s, 10).to_base(16))

Now, either my logic is wrong or I have executed it wrong, or it is correct and I just don't know it. The difficulty is, I don't know what the correct answer is, and it's hard to check, because there aren't too many float converters out there (lots of integer ones though).

[OneOffDriveByPoster]

50.05 in hex = ?

50 in hex: 32

.05 in hex = ?

5 / 100
5h / 64h

Long division

code:

0.0CCC... 64    5 00         500

You are looking for 32.0CCC...

[Tyr_God_Of_War]

You know, the calculator that comes with windows can't do that. It just rounds. That ruins my cunning plan of using it to check your work. Maybe another, better calculator can...

[Insectoid]

Well, I'm not on windows, and so don't have access the that calculator, even if it did work. The calculator that comes with mac is worse than the windows one.

[OneOffDriveByPoster]

Tyr_God_Of_War @ Fri Dec 19, 2008 9:03 pm wrote:

You know, the calculator that comes with windows can't do that. It just rounds. That ruins my cunning plan of using it to check your work. Maybe another, better calculator can...

Try

320CCCCCCCCCCCCC
/100000000000000

in decimal:

3606482581598293196 / 72057594037927936
= 50.049999999999999988897769753748 approx.

(This was done with the Windows calculator.)

[Insectoid]

So, I would have to divide the number by the base equivalent of 100? so to octal would be 5/144, and to binary would be 101/1100100. Right?

[OneOffDriveByPoster]

Well 0.05 = 5/100.

You could do 1/20...

Converting from hex or oct to binary and vice versa is much simpler of course.

8 = 2^3 so an oct digit is 3 bits
16 = 2^4 so a hex digit is 4 bits

Here is the binary case.

code:

v   v            0.000011001 1100100  101 00000  <==           11 00100            1 111000            1 100100               10100000  <==                1100100

[Insectoid]

I just though of something...

In base 10, you can multiply a decimal by 10 multiple times to make it >1. So, for hex, could I multiply by 16^length to make it >0, convert, then divide by 16^length again? Just a thought...

[OneOffDriveByPoster]

Exactly:

If you consider using multiplication, which the division process
implies when you "add zeros".

For each digit after the ones digit, you can multiply by the base to see
what the value of the digit is by taking the non-fractional part.

This is obvious using base-10.

code:

0 . 0| 0.05 * 10 = 0.5 5| 0.5 * 10 = 5

0.05 = 0.05 (as expected)

code:

0 . 0| 0.05 * 2 = 0.1 0| 0.1 * 2 = 0.2 0| 0.2 * 2 = 0.4  <== 0| 0.4 * 2 = 0.8 1| 0.8 * 2 = 1.6 1| 0.6 * 2 = 1.2 0| 0.2 * 2 = 0.4  <==

0.05 = 0.0000110011001100...
(This is what we got from long division too.)

code:

0 . 0| 0.05 * 16 = 0.8 C| 0.8 * 16 = 12.8  <== C| 0.8 * 16 = 12.8  <==

0.05 = 0.0CCC...
(This is what we got from long division too.)

[OneOffDriveByPoster]

insectoid @ Wed Dec 24, 2008 12:08 pm wrote:

I just though of something...

In base 10, you can multiply a decimal by 10 multiple times to make it >1. So, for hex, could I multiply by 16^length to make it >0, convert, then divide by 16^length again? Just a thought...

You need to keep the remainder and deal with it. The multiplying process above is basically this.

[Insectoid]

I got it! Multiply the number by target base^number of decimal places, divide by target base equivalent of 100 (64, 144, etc.), and then divide by base^number of decimal places. I think there's something missing, but I'm not sure...