From: Harris Reynolds <hreynolds2@...>
Date: 2006-06-09T11:39:09+09:00
Subject: Re: Converting 8 bytes to a Float

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Mike... thanks for the tip.  My mileage is varying though... here is the same irb session on my machine:
 
irb(main):004:0> s = ""
=> ""
irb(main):005:0> s << 64 << 53 << "\0" * 6
=> "@5\000\000\000\000\000\000"
irb(main):006:0> s.unpack('D')
=> [6.73510288410787e-320]
 
I'm not getting the clean result for some reason.
 
~harris

----- Original Message ----
From: Mike Stok <mike@stok.ca>
To: ruby-talk ML <ruby-talk@ruby-lang.org>
Sent: Thursday, June 8, 2006 8:57:06 PM
Subject: Re: Converting 8 bytes to a Float


On 8-Jun-06, at 8:47 PM, Harris Reynolds wrote:

> I am still not getting the results I am looking for hacking around  
> with the couple suggestions I received.  Here is an example:
>
> bytes = Array.new(8)
> bytes[0] = 64
> bytes[1] = 53
> bytes[2] = 0
> bytes[3] = 0
> bytes[4] = 0
> bytes[5] = 0
> bytes[6] = 0
> bytes[7] = 0
>
> Calling:
>
> binary_string = bytes.pack('D')
>
> returns a binary string, but not the float which should be 21.0; if  
> I turn around and call
>
> binary_string.unpack('d'), I end up with the original bytes.
>
> It is turning out that getting from bytes to a Float object is a  
> bit painful.  Is there something easy I am missing?

You should be using unpack ;-) .  In irb:

ratdog:~ mike$ irb
irb(main):001:0> s = ""
=> ""
irb(main):002:0> s << 64 << 53 <<"\0" * 6
=> "@5\000\000\000\000\000\000"
irb(main):003:0> s.unpack('D')
=> [21.0]

Mike

-- 

Mike Stok <mike@stok.ca>
http://www.stok.ca/~mike/

The "`Stok' disclaimers" apply.
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