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?
Program 1976
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07-05-2006, 02:38 PM
RCL 1 R9 = ((R1+R2)*R3)+R4 Is this how it works?
07-05-2006, 02:45 PM
R9=((R3+R4)*R2)+R1 Edited: 5 July 2006, 2:50 p.m.
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 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...
07-05-2006, 02:55 PM
Thank you!
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. |
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