On Mon, 16 Jan 2023 14:25:04 +0200, Binyamin Dissen wrote: >The only way I could figure out to do it was : > > LCLA &I,&J >&J SETA exponent (an EQU) >&I SETA (2 SLA &J-1) >value EQU &I > >Is there no better way? > Do you need only powers of 2?
What's the precedence of operators? (2 SLA (&J-1)) or ((2 SLA &J)-1) Why not simply: &I SETA (1 SLA &J) ? (I feel old. I remember when conditional and regular operations such as EQ and SETA couldn't be mixed.) -- gil ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [email protected] with the message: INFO IBM-MAIN
