Program 1976 mountain Junior Member Posts: 5 Threads: 2 Joined: Jan 1970 07-05-2006, 02:38 PM RCL 1 RCL 2 RCL 3 RCL 4 + x + STO 9 R9 = ((R1+R2)*R3)+R4 Is this how it works? Etienne Victoria Senior Member Posts: 464 Threads: 10 Joined: Jan 2005 07-05-2006, 02:45 PM R9=((R3+R4)*R2)+R1 Edited: 5 July 2006, 2:50 p.m. Kiyoshi Akima Senior Member Posts: 325 Threads: 18 Joined: Jul 2006 07-05-2006, 02:49 PM No, you get R9=((R3+R4)*R2)+R1 A stack diagram may make this clearer (top of stack to the right, X to the left): ```Op X Y Z T RCL 1 R1 RCL 2 R2 R1 RCL 3 R3 R2 R1 RCL 4 R4 R3 R2 R1 + (R3+R4) R2 R1 * R2*(R3+R4) R1 + R1+(R2*(R3+R4)) ``` Each RCL puts a new item on the stack, pushing up the old contents. The operators work on the bottom one or two items. Hope this didn't confuse you even more... mountain Junior Member Posts: 5 Threads: 2 Joined: Jan 1970 07-05-2006, 02:55 PM Thank you! mountain Junior Member Posts: 5 Threads: 2 Joined: Jan 1970 07-05-2006, 03:03 PM I'm trying to decipher a 1976 HP program and I find it a little confusing. In 1976, I was writing programs for the TI equivalent. But the diagram of the stack helps a lot. Thank you. « Next Oldest | Next Newest »

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