From: "Peter Relson" <rel...@us.ibm.com>
Sent: Wednesday, April 27, 2022 9:30 AM
Apologies if this was mentioned, since I am only now getting around to reading some of this
thread:
Bernd wrote (and others had similar)
LPR: if the register contains 0x80000000, IMO the result will be zero
(and overflow),
so you're right ... this will lead to a zero result. IMO, the overflow
will be ignored.
If the mask bit is zero, there will be no interrupt. The overflow CC will be
zset.
If the mask bit is 1, an overflow interrupt will occur.
The question is about "will be ignored". Do you know that it will truly be
ignored?
See above. If the mask bit is 1, the interrupt will occur.
Can you be certain that the program mask bit is not on which, if on, would cause the overflow to
program-check?
If running under LE with ESPIE protection, might that program mask be on?
z/OS itself cannot make such an assumption in many paths (in an SVC path,
it might be able to because the SVC new PSW has 0's for the program mask;
in a PC path, it cannot because the PC instruction does not change the
program mask so the routine would have to do so). Thus this nice trick
to avoid compares and branches is not, in general, usable within the OS.
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