Initializing Values in a Program (HP35s)

[Chuck]

Suppose you need to initialize X=2 and Y=3 in a program. One method is:

001  2

002  STO X

003  3

004  STO Y

A second method is to use EQN before each line

001  2 -> X

002  3 -> Y

where -> is the STO command.

The second method obviously reduces the number of lines of code, but is one method any more efficient (timewise) over the other? I haven't been able to discern a difference in my small programs, so it's just really a matter of curiousity.

CHUCK

[Egan Ford]

Create a loop and time it:

E.g.:

A001 100

A002 STO I

A003 your code here

...

An   DSE I

An+1 GTO A003

An+2 RTN

This will run your code 100 times. Use a stopwatch to time it.

Edited: 3 Oct 2007, 1:03 p.m.

[Arne Halvorsen (Norway)]

Y001 LBL Y

Y002 0.999

Y003 STO X

Y004 2

Y005 STO Y

Y006 ISG X

Y007 GTO Y004

Y008 RTN

Used 1m 8s

Y001 LBL Y

Y002 0.999

Y003 STO X

Y004 eq 2>Y

Y005 ISG X

Y006 GTO Y004

Y007 RTN

Used 1m 57s

Edited: 3 Oct 2007, 1:22 p.m.

[Chuck]

Okay, I looped it 100 times.

11 seconds with

2

sto x

3

sto y

and 20 seconds with

2 -> x

3 -> y

Looks like there is definitely a trade-off with speed -vs- size.

[Meenzer]

The difference in length is only

LBL A

2

STO


LN=67



LBL A

2>A

LN=66

Edited: 3 Oct 2007, 2:43 p.m.

[Arne Halvorsen (Norway)]

Equation are parsed run time every time used, so clearly there will by an overhead.

[Paul Dale]

I've not checked if this is the case here or not but don't trust the reported program lengths, they are wrong.

To measure sizes more accurately, try looking at the change in the bytes free in the memory menu.

- Pauli