In article <[email protected]> you wrote: > You forgot to specify whether the top 32 bits of R0 and R15 must be zero > after the lower 32 bits have been manipulated. > Assuming that you do, then:
> LLGFR 15,3 /* Load bits 32-63 in R3 into bits 32-63 of R15 and > clear bits 0-31 of R15 */ > XGR 0,0 /* Initialize 64-bit R0 to zero */ > SLLG R0,R3,32 /* Shift bits 0-31 in R3 to bits 32-63 in R0 */ > The original contents of R3 is preserved and the above can be > executed in any AMODE > Best regards, Andre One caution. If the contents of the lower half of R3 are meant to be used to address memory below the bar, the contents of bit 32 need to be dealt with. In this case, LLGTR would be the correct instruction. e.g. LLGTR 15,3 /* Load bits 33-63 in R3 into bits 33-63 of R15 and clear bits 0-32 of R15 */ -- Don Poitras - SAS Development - SAS Institute Inc. - SAS Campus Drive [email protected] (919) 531-5637 Cary, NC 27513 ~ ~ ~ ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [email protected] with the message: INFO IBM-MAIN
