array := array

codeking

Does the assignment operator (:=) work with arrays.· For example, we have array1 and array2.· Each one has 12 bytes.· If we do array1 := array2, will array 2's data be copied to array1?· If not, how do I do this?

Mike Green

The assignment operator does not work with arrays. You can use bytemove/wordmove/longmove to do a copy (look at the manual) or you can do a loop with something like "repeat i from 0 to 11", then "array1[noparse][[/noparse] i] := array2[noparse][[/noparse] i]".

Bergamot

I think the assignment operator would cause both array variables to point to the same array in memory (which does not sound like what you want).

Instead, I'd use

bytemove(@Array1, @Array2, 12)

EDIT: Curse you Mike Green!

EDIT #2: Is it faster to loop by hand? Is longmove faster than bytemove or wordmove?

Mike Green

Bergamot,
If the array is word or long aligned, wordmove is faster and longmove is fastest since 2 bytes or 4 bytes are moved with two Hub accesses.
Mike

codeking

is there some way I could do something like this?

IF array1 == array2

Mike Green

None of the operators in Spin work on arrays. You could write a subroutine (method) that takes two array addresses and compares them byte by byte and returns either true or false like:

PRI compare(addr1,addr2,size)
   repeat size
      if byte[noparse][[/noparse]addr1++] <> byte[noparse][[/noparse]addr2++]
         return false
   return true

Then you could write "if compare(@array1,@array2,12)"

codeking

Thanks, Mike

asterick

Mike Green said...
Bergamot,
If the array is word or long aligned, wordmove is faster and longmove is fastest since 2 bytes or 4 bytes are moved with two Hub accesses.
Mike

Do RD{WORD|LONG} require that they be on their appropriate boundaries? the docs don't say anything about that.

Bergamot

asterick said...

Mike Green said...
Bergamot,
If the array is word or long aligned, wordmove is faster and longmove is fastest since 2 bytes or 4 bytes are moved with two Hub accesses.
Mike

Do RD{WORD|LONG} require that they be on their appropriate boundaries? the docs don't say anything about that.

I think he meant that it's faster if the words/longs are properly aligned, but works (though more slowly) if they are not.

Regardless, the docs don't seem to address that issue either.

Paul Baker

Any WORD|LONG operation works on aligned boundries only, the Propeller masks out bits of the address to ensure this (last LSB for WORD, last two LSB for LONG).

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Paul Baker
Propeller Applications Engineer

Parallax, Inc.

asterick

Good to know. *scampers off to fix a bug in gear*