O changes the register, not the storage operand, so if you want to do it this way, you would have to do O, followed by ST.
(same goes for N, X). similar to A, S, interesting problems with M and D Kind regards Bernd Am 21.05.2021 um 00:30 schrieb Charles Mills:
No, I mean O into the target field; O into the appropriate byte of the bit array. O no longer requires alignment, right? So you just set the appropriate bit in positions 0-7 of the register and zero the remaining positions and O it against the target field. There is a three-byte "hangover." It won't hurt anything unless you happen to be at the end of a protection key or addressability boundary. That is what I meant by "if one could tolerate the storage reference 3 bytes beyond ..." It would be easy enough in many cases to assure three bytes of padding. No bits would ever be altered in those 3 bytes, but the hardware would see a storage reference. I suppose in the boundary case it could cause a cache miss. Charles
