Thanks

-----Original Message-----
From: IBM Mainframe Discussion List [mailto:[EMAIL PROTECTED] On
Behalf Of (IBM Mainframe Discussion List)
Sent: 08 November 2007 04:20 nm
To: [email protected]
Subject: Re: CSA 'above the bar'


 
 
In a message dated 11/8/2007 10:02:17 A.M. Central Standard Time,  
[EMAIL PROTECTED] writes:

I have  not checked this for myself yet, and probably won't have the
time
in the  next few weeks... In theory... If I allocate 200 bytes of
storage
at  x'7fffff00', and my program, the way some Cobol programs are, writes
1600  bytes there, what would the addresses be, would it jump across the
'hole'  or try and write over the hole?

The addresses of the 200 bytes (decimal number 200 is assumed) that your

program allocates would be X'7FFFFF00' to X'7FFFFFC7'.  If your COBOL
program 
attempts to write 1600 (decimal assumed again) bytes beginning at
X'7FFFFF00', 
your program would write at most the first 256 of the 1600 bytes  and,
when 
attempting to write the 257th byte, would be interrupted with a
protection 
check program interrupt that would result in a S0C4 ABEND.  I  said "at
most" 
because I have not yet studied the latest PoOps to know how new  move
instructions 
move bytes.  And it would depend on how your program does  the move.  If
it 
moves one byte per instruction in a loop, it would write  256 bytes and
then 
ABEND.  There might be another type of move  instruction whose
preprocessing 
checks the beginning and ending addresses  for validity before moving
the first 
byte, in which case your program would  write zero bytes and then ABEND.
Your 
program would try to write over the  hole but would not be allowed to by
the 
various protection mechanisms in the  processor architecture.

 
Bill  Fairchild
Franklin, TN



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