Surely not.  It checks 16 bytes.

The first 'iteration' compares offset 1 to 0; the 2nd, 2 with 1, and so on, 
until the 15th 'compare' checks offset 15 with offset 14.

In total, offset 0 would be compared (for equality) with all bytes up to offset 
15. 0 .. 15 = 16.

Trivially, CLC  1(1,R5),0(R5) compares whether the two bytes are the same.

> -----Original Message-----
> From: [email protected]
> Sent: Thu, 30 Sep 2010 15:35:49 -0400
> 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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