I don't think you are correct. This is an overlap compare, it compares
+0 to +1 for 15 bytes. Example
0 1 2 3 4 5 6 7 8 9 A B C D E F
1 X X
2 X X
3 X X
4 X X
5 X X
6 X X
7 X X
8 X X
9 X X
10 X X
11 X X
12 X X
13 X X
14 X X
15 X X
0 1 2 3 4 5 6 7 8 9 A B C D E F
As you can see, it compares all 16 bytes (since the comparer is offset
by 1 byte).
Frank M. Ramaekers Jr.
-----Original Message-----
From: IBM Mainframe Assembler List
[mailto:[email protected]] On Behalf Of Watkins, Douglas
Sent: Thursday, September 30, 2010 2:36 PM
To: [email protected]
Subject: Re: Help jog my memory (16-bytes the same)
This statement checks only 15 bytes, not 16 bytes (starting at the
second byte pointed to by R5). (And, higher and lower, as well as same,
may be indicated.)
-----Original Message-----
From: IBM Mainframe Assembler List
[mailto:[email protected]] On Behalf Of Mike Flint
Sent: Thursday, September 30, 2010 3:29 PM
To: [email protected]
Subject: Re: Help jog my memory (16-bytes the same)
Yes, it does.
Although I'd apply the appropriate commentary in most circumstances (as
a memory jogger).
Mike.
> -----Original Message-----
> From: [email protected]
> Sent: Thu, 30 Sep 2010 14:14:10 -0500
> To: [email protected]
> Subject: Help jog my memory (16-bytes the same)
>
> Doesn't the following statement check to see that all 16-bytes are
same?
>
>
>
> 00251A D50E 5001 5000 00001 00000 8688 CLC 1(15,R5),0(R5)
>
>
>
>
>
>
>
> Frank M. Ramaekers Jr.
>
>
>
> Systems Programmer
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>
>
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