Thanks to all! Especially pointing out different ways to do it, like this.
The first assembler language I learned to any proficiency was for a PIC microcontroller. (RISC and Harvard) It had I believe only 33 instructions which were easy to learn. On the other hand, you really had to think to make those few do all the things you wanted to do. That I liked. This on the other hand makes me go Doh! (-: I learned a shift left instruction without thinking there was an equivalent to shift to the right as well. I totally LOL'd at myself on this one. On the other hand the amount of and complexity of (some of) the zArch instructions can be overwhelming to a new learner. I couldn't do it without you guys. (-; Thanks. (ps. I accidentally posted my original to IBM-MAIN while thinking I was posting to the Assembler list.) -----Original Message----- From: IBM Mainframe Discussion List on behalf of Bill Wilkie Sent: Fri 12/28/2007 4:00 PM To: IBM-MAIN@BAMA.UA.EDU Subject: Re: Shifting nibbles to the end of a register Lindy: You mention moving a nibble but your INTENDED code doesn't do that. You are just moving the 9A from the LO byte to the HO byte. A nibble as I recall is 4 bits of a byte, either the LO or HO 4. To move out a nibble: LA R4,X'9A' SRDL R4,4 SRL R5,28 That will yield: R4= 00000009 R5= 0000000A Alternatively, to move it to the HO byte of R5 instead of to the LO byte, SRL R5,4. You could then do a STCM R5 of either the HO byte or the LO byte. Using the algebraic Shift involves a sign so use the logical. Bill ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [EMAIL PROTECTED] with the message: GET IBM-MAIN INFO Search the archives at http://bama.ua.edu/archives/ibm-main.html
