Graeme,

Sorry, but I just have to do this.  X86 registers work in trinary mode rather 
than binary?  Your map of the eax register shows.

eax:   00000000 00000000 00000001 00000020   <======  What's the 00000020 mean? 
 :-)

Rex

-----Original Message-----
From: IBM Mainframe Discussion List [mailto:[email protected]] On Behalf Of 
Graeme Gibson
Sent: Monday, April 11, 2011 11:42 PM
To: [email protected]
Subject: Re: New job for mainframes: Cloud platform

Well, let's not skew the kiddie's brains too much..

"Endian- ness", in the context used in these posts, refers to BYTE
order not BIT order.

The bit-order *within each byte* is still "most significant on the
left" in both big- and little- endian systems.

Another wrinkle:  In the x86 world (little-endian) Intel "names" the
bits within a byte, from the left, 7,6,5,4,3,2,1,0 whereas in the z..
world (big-endian) IBM "names" them 0,1,2,3,4,5,6,7.  Regardless of
the naming scheme used, the "rightmost" bit in each byte is the
least-significant. :-)

An x86 code fragment might help to illustrate:

         .code
         mov eax,258     ;let's start with 258 KG
         mov weight,eax
         ..
         ..
         ret
         .data
weight  dd  0           ;weight in kilograms

After that 2nd mov instruction, while the 32-bit eax register looks like:

        bits 32,31,..,24
        |        bits 23,22,..,16
        |        |        bits 15,14,..,8
        |        |        |        bits 7,6,5,..,0
        |        |        |        |
eax:   00000000 00000000 00000001 00000020

..the four bytes at label "weight" will look like:

weight 00000010 00000001 00000000 00000000
        |        |        |        |
        |        |        |        bits 32,31,..,24
        |        |        bits 23,22,..,16
        |        bits 15,14,..,8
        bits 7,6,5,..,0

So, bit numbering aside, it still looks like a big-endian world once
you're "inside" the processor.

Cheers,
Graeme

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