How Would you count a greater value than 65535 Please give a very simple examp

sam_sam_sam

·I know that this question is very simple for most of you but..........

I·know that with P Basic you would

do this to count

count = count + 1

and you only count 65535 and then it would roll over

and have· another counter start when the first one got to 65534 then start your next counter

The only thing that I have done with the Propeller is the few being example when this chip first came out I would like to learn more about programing this chip as well



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Sam

BradC

sam_sam_sam said...
I know that with P Basic you would



do this to count



count = count + 1



and you only count 65535 and then it would roll over

The propeller won't roll over until you get to 4294967295 as it counts in 32 bits natively rather than 16. Just declare your counter variables as LONG rather than WORD or BYTE.

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lt's not particularly silly, is it?

Ariba

With SPIN you would do this to count:

count := count + 1

or this:

count += 1

or this:

count++

or this:

++count

All this variations increment the variable count by 1. If count is a long-variable (=32 Bit) , then you can count up to 2_147_483_647, then the value rolls over to -2_147_483_648 (That is because SPIN handles longs as signed variables).

If you need to count higher then use more than 1 variable and count the roll-overs in a second count variable:

count1 := count1 + 1
  if count1 == 2_147_483_647
    count2 := count2 + 1
    count1 := 0

this can also be written like this:

if ++count1 == POSX
     count2++
     count1~

Andy

James Newman

If you need more than that, ASM add and subract routines can use the overflow bit from a previous operation, allowing you to chain them together for a value as big as you need. Look around the forum, I remember a post about 64 bit numbers that demonstrated it.

localroger

There's actually a more economical way to do multiword count in Spin:

if not (++count)
counthiword ++

This works for both signed and unsigned numbers, as long as you treat counthiword as the 2's complement extension of count. This works by (1) incrementing count, (2) logical NOT converts any nonzero value to zero and only zero to -1, and if it's -1 (3) carries to the hiword.

mpark

Possibly slightly more economical:

ifnot ++count
  counthiword++

Saves you a not operation.

localroger

@mark, I totally forgot about ifnot. Spin is deeper then Loch Ness, I tells ya.

sam_sam_sam

I want to thank every one that replayed to this post

Arida

Can·you help little more with this in explain what is the different between each one that you have here in the way that it would count

count := count + 1

or this:

count += 1

or this:

count++

or this:

++count

or this

·localroger and mpark

··············· if not (++count)
··············· counthiword ++

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Sam

Post Edited (sam_sam_sam) : 8/31/2009 11:54:10 PM GMT

localroger

@sam: As standalone statements, all of the code-bracketed examples you give have exactly the same effect. The only difference is with count++ vs. ++count, if you use that as the ARGUMENT for another function (such as p := count++ or p := ++count) count++ returns the value BEFORE the ++ increment, while ++count returns the value AFTER the ++ increment. Both leave count one higher after the operation.

What the ifnot thingie does is it checks to see if the result of the ++count is zero, which means it has rolled over; just as if you are adding in a one-digit decimal register and you get 7, 8, 9, 0 ... that 0 means there should be a 1 added to the tens. So by checking ++count for zero we are asking, did we just roll over? And if the answer is yes, we add 1 to the next higher "digit." That gives you 64 bit counts (and more if you extend the idea). And that can be a useful thing when the CNT register overflows after 2 minutes or so at 80 MHz.

sam_sam_sam

localroger

Thank you for exlplaine this to me that helps a Lot·

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Sam

Ole Man Earl

get more fingers !