If... or... but with a twist

[Roboteernat]

Is it possible somehow to...

make an if or statement where the variable its comparing to is plus 10 or negative 10, ie

<code><pre><font size=2 face='Courier'>if b1 &lt;= (b12 - 10) or b1 &gt;= (b12 +10) then ... </font></pre></code>

the command goes to ... if the value for b1, say 100 falls not between b12, say 200, does not fall between 190 and 210.

or would i need to use 2 registers to save the b12 - 10 and b12 +10 values.

thanks for any help

nat poate
as i need to use all the other registers and only have b12 left.

 

[inglewoodpete]

You can't do your mathematics in the IF command. The IF statement can only handle simple comparisons. (Eg If A = B and C &gt; D Then)

Have a look at page 34 of the command manual. I seem to remember that this was described in detail somewhere else in the manual(?)

 

[Roboteernat]

okies, it was just a hope, i think i have arranged my code now so that i can use some other registers, lets hope it still works! wooo

thanks anyways.

ps close, twas 27 and 28

nat poate

 

[tarzan]

On Picaxes&#8217; that support infrain, infrain2 or keyin, will give you an extra variable if not in use.

Namely: Infra/Keyvalue

Except 08M which uses B13 for Infrain2.

08M infra = another term for variable b13, used within the 08M infrain2 command
18 18A 18X 28A 28X 40X infra = a separate variable used within the infrain command
keyvalue = another name for infra, used within the keyin command

Edited by - tarzan on 07/02/2007 04:15:03

 

[demonicpicaxeguy]

if b1 &lt;= (b12 - 10) or b1 &gt;= (b12 +10) then


below should work unless you have no more variables left

let b11 = b12 - 10
if b1 &lt;= b11 then goto examplesub1
let b11 = b12 + 10
if b1 &gt;= b11 then goto examplesub2

 

[Jeremy Leach]

Just for general interest since it looks like you've solved the problem, here's (untested) code that just might help, but needs a bit of explaining!...

<code><pre><font size=2 face='Courier'>
b1 = b1 - 10 MIN b12 - b12 MAX 1 XOR 1 * b1 + 10 Max 12
If b1 &lt; b12 then Goto OutsideRange
</font></pre></code>

Explaining the calculation in sections...
A: <i> b1 - 10 MIN b12 </i>
If b1 is greater than (b12 + 10) then the result of this expression will be &gt;b12

B: <i> - b12 MAX 1 </i>
This will then transform the result to 0 or 1. 1 means b1 &gt;(b12 + 10) and 0 means b1 &lt;= (b12 + 10).

C: <i>XOR 1 </i>
This reverses the meaning of the 1 and 0, so now 0 means b1 &gt;(b12 + 10) and 1 means b1 &lt;= (b12 + 10).

D: <i>* b1 </i>
This gives a lead in to the final calculation. If b1 &gt;(b12 + 10) then it will give 0, and if b1 &lt;= (b12 + 10) it will give b1.

E: <i> + 10 Max 12 </i>
This is the opposite of the first part of the calculation, shown in A, and testing for if b1 &lt; (b12 - 10). However because of step D it also is a final test for if B1 is outside the ranges. If the final result &lt;b12 then it is outside.

I think !