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RCL 1
RCL 2
RCL 3
RCL 4
+
x
+
STO 9

R9 = ((R1+R2)*R3)+R4

Is this how it works?

R9=((R3+R4)*R2)+R1

Edited: 5 July 2006, 2:50 p.m.

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...

Thank you!

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.