By definition, base and index registers are treated as 64 bit binary values (unsigned), with only the relevant bits (24, 31, or 64) used. The relevant bits are simply added with overflow discarded. There is no sign bit to ignore.
In Chapter 3 (Storage), section Address Wraparound: When, during the generation of the address, an address is obtained that exceeds the value allowed for the address size (2^24 - 1, 2^31 - 1, or 2^64 - 1), one of the following two actions is taken: 1. The carry out of the high-order bit position of the address is ignored. This handling of an address of excessive size is called wraparound. 2. An interruption condition is recognized. ... Addresses generated by the CPU that may be virtual addresses always wrap. Eric Rossman, CISSPĀ® ICSF Cryptographic Security Development z/OS Enabling Technologies [email protected] On Sat, 23 Oct 2021 11:33:33 -0500 Joe Monk <[email protected]> wrote: > Howdy, > > I know this will probably be an easy answer for somebody... but I dont deal > with AM64 much. > > If Im in AM64 and I load an index register with -1, does the machine ignore > the sign when using it in an RX instruction such as STC? > > I know it ignores the sign in AM24/31... ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [email protected] with the message: INFO IBM-MAIN
