Bit manipulation???

[James Long]

Ok guys....have a question....not really difficult...but I would like to know.

Is it possible to manipulate one bit within the propeller? Of the variable type.....not on the registers or pins.

Please bear with me....I'm a ladder logic person....not really a written code type.

I would like to use bits as variables/flags....but I don't see a way to use just one bit. Using a byte is just huge overhead.

Thanks,

James L

[Mike Green]

All bit manipulation on the propeller is done with 32 bit integer logical operations (&, |, ^, !) including the many different kinds of shift operations. For using individual bits as flags, use a mask like $00400000, preferably as a named constant (for clarity) and use the bit logical operations so to set the bit do "x |= $00400000", to clear the bit do "x &= !$00400000", to test if set do "(x & $00400000) <> 0", and to test if clear do "(x & $00400000) == 0". If you're careful, you can do combinations of bits the same way.

[cgracey]

If you have over 32··bits, you·could make an array·of longs with routines to read and write by bit number:

VAR long Bits[32] '1024 bits

PRI ReadBit(BitNum) : Bit
· Bit := Bits[BitNum >> 5] >> BitNum & 1

PRI WriteBit(BitNum, State)
· Bits[BitNum >> 5] |=·|< BitNum
· Bits[BitNum >> 5] ^=·|< BitNum &·not·State·

If you have only 32 bits, you could use the above code, but get rid of the [BitNum >> 5] indexes:

VAR long Bits '32 bits
·
PRI ReadBit(BitNum) : Bit
· Bit := Bits >> BitNum & 1
·
PRI WriteBit(BitNum, State)
· Bits |=·|< BitNum
· Bits ^=·|< BitNum &·not·State·
·


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Chip Gracey
Parallax, Inc.

[James Long]

Mike said...
All bit manipulation on the propeller is done with 32 bit integer logical operations (&, |, ^, !) including the many different kinds of shift operations. For using individual bits as flags, use a mask like $00400000, preferably as a named constant (for clarity) and use the bit logical operations so to set the bit do "x |= $00400000", to clear the bit do "x &= !$00400000", to test if set do "(x & $00400000) <> 0", and to test if clear do "(x & $00400000) == 0". If you're careful, you can do combinations of bits the same way.
Mike....the mask has me confused. Setting them.....Ok.....I get that part. I was going to do· combinations.....that way I could have 32 flags in one group.


Chip said...
If you have over 32··bits, you·could make an array·of longs with routines to read and write by bit number:

VAR long Bits[32] '1024 bits

PRI ReadBit(BitNum) : Bit
· Bit := Bits[BitNum >> 5] >> BitNum & 1

PRI WriteBit(BitNum, State)
· Bits[BitNum >> 5] |=·|< BitNum
· Bits[BitNum >> 5] ^=·|< BitNum &·not·State·

If you have only 32 bits, you could use the above code, but get rid of the [BitNum >> 5] indexes:

VAR long Bits '32 bits
·
PRI ReadBit(BitNum) : Bit
· Bit := Bits >> BitNum & 1
·
PRI WriteBit(BitNum, State)
· Bits |=·|< BitNum
· Bits ^=·|< BitNum &·not·State·
·

Chip I know I could have over 32 bit as flags.....but really doubtful. I use them alot ...but not in huge amounts.

I'll have to read up on your method of affecting them.....but I think I get it.

James

[Mike Green]

James,
If you'd prefer using a bit number, you can use the "|<" operator. Setting Bit 22 is "x |= |<22". Clearing Bit 22 is "x &= !|<22". Testing Bit 22 to see if it's set is "(x & |<22) <> 0". Toggling Bit 22 is "x ^= |<22" and so on.
Mike

[Newzed]

Mike, if you wanted to toggle a bit, couldn't you write:

x := !|<22

Sid

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Sid Weaver
Don't have VGA?

Newzed@aol.com (mailto:Module?Newzed@aol.com)
·

[cgracey]

You could even use some innocuous register like VSCL as a bit-addressable flag buffer:

Bit := VSCL[BitNum]···· 'read a bit
VSCL[BitNum] := Bit···· 'write a bit
!VSCL[BitNum]···········'toggle a bit

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Chip Gracey
Parallax, Inc.

Post Edited (Chip Gracey (Parallax)) : 10/20/2006 6:22:52 PM GMT

[James Long]

Chip,

Now that was very helpful.......Thanks.



Mike,

Thanks for the suggestion.



James L

[Mike Green]

Sid,
"x := !|<22" doesn't mean what you think it does. The "!" is a logical "NOT" operator so "|<22" is $00400000 and "!|<22" is $FFBFFFFF, so your statement sets x to $FFBFFFFF.
Mike

[cgracey]

Mike,
You're right. It would need to be:
x ^= |<22········ 'xor $00400000 into x


Mike Green said...
Sid,
"x := !|<22" doesn't mean what you think it does. The "!" is a logical "NOT" operator so "|<22" is $00400000 and "!|<22" is $FFBFFFFF, so your statement sets x to $FFBFFFFF.
Mike

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Chip Gracey
Parallax, Inc.