Dectect morethan 1 bit set in a byte.

Darbs003 · Aug 6, 2010

Hi there,

Can any tell me a simple way of decting if more than one bit is set in a byte.
when using a S7 400.

Cheers

erdemsvri · Aug 6, 2010

L MB 10
L B#16#0
==I
JCN a1
S M 1.2
JU a2
a1: R M 1.2
a2: NOP 0

I think this code will solve your problem.

M1.2 = 1 if MB10 == 0
M1.2 = 0 if mb10 <> 0

L D[AR2P#0.0] · Aug 6, 2010

You are detecting more than zero bits set - the op wants to know if more than 1 bit is set.

Darbs003 · Aug 6, 2010

Need to dectect when more than one bit position is high, not when the value of the byte is not zero. any more ideas?

KalleOlsen · Aug 6, 2010

I can't think of any other solution that a shift loop. When the thrown bit = 1, check if the rest value <>0.
Kalle

L D[AR2P#0.0] · Aug 6, 2010

One implementation:-

Code:

FUNCTION FC 5 : BOOL
TITLE =
VERSION : 0.1

VAR_INPUT
  byData : BYTE ; 
END_VAR
VAR_TEMP
  b1Bitset : BOOL ; 
END_VAR
BEGIN
NETWORK
TITLE =
      SET   ; 
      R     #b1Bitset; 
      R     #RET_VAL; 
      L     #byData; 
      SRW   1; //bit 0
      JZ    S1; 
      SET   ; 
      S     #b1Bitset; 
S1:   SRW   1; //bit 1
      JZ    S2; 
      A     #b1Bitset; 
      S     #RET_VAL; 
      BEC   ; 
      S     #b1Bitset; 
S2:   SRW   1; //bit 2
      JZ    S3; 
      A     #b1Bitset; 
      S     #RET_VAL; 
      BEC   ; 
      S     #b1Bitset; 
S3:   SRW   1; //bit 3
      JZ    S4; 
      A     #b1Bitset; 
      S     #RET_VAL; 
      BEC   ; 
      S     #b1Bitset; 
S4:   SRW   1; //bit 4
      JZ    S5; 
      A     #b1Bitset; 
      S     #RET_VAL; 
      BEC   ; 
      S     #b1Bitset; 
S5:   SRW   1; //bit 5
      JZ    S6; 
      A     #b1Bitset; 
      S     #RET_VAL; 
      BEC   ; 
      S     #b1Bitset; 
S6:   SRW   1; //bit 6
      JZ    S7; 
      A     #b1Bitset; 
      S     #RET_VAL; 
      BEC   ; 
      S     #b1Bitset; 
S7:   SRW   1; //bit 7
      JZ    S8; 
      A     #b1Bitset; 
      S     #RET_VAL; 
      BEC   ; 
S8:   BEU   ; 
END_FUNCTION

bernie_carlton · Aug 6, 2010

I don't have the PLC specific code but

If byte = 0 then exit false

byte = byte AND (byte - 1) - this clears the least significant bit

If byte = 0 then exit false else exit true

For reference see 'Bit Twiddling Hacks'

TurpoUrpo · Aug 6, 2010

Yeah, that is nice.

Code:

function fc5 : bool

var_input
    byIn : byte;
end_var

begin

l 0;
l byIn;
==I;
bec;
l -1;
+I;
aw;
l 0;
==I;
bec;
set;
s ret_val;

end_function;

For s7-400 there is some optimization to do, i just dont right now remember how four accus behave.

tvey · Aug 6, 2010

You can use logarithms to determine if one bit is set:

Code:

Log(x) / Log(2)

If the result has a non-zero decimal portion, more than one bit is set. If the result is an integer, only one bit is set.

Let's say you have a word value of 128:

Code:

Log(128) / Log(2) = 7.0

There is no remainder, so only one bit (Bit 7) is set.

Let's say you have a word value of 129:

Code:

Log(129) / Log(2) = 7.01122

There is a remainder of 0.01122, therefore, more than one bit is set. In this case, Bit 7 and Bit 0.

To implement this logic in a PLC:

Compute the logarithm and store the result in a floating point type register (REAL, etc).
Copy the result to an Integer register. In most PLC's this will round or truncate any decimal portion in the float.
Compare the float to the integer. Most PLCs will convert the integer to a float for the comparison. If the floating point result has a non-zero decimal portion, the compare will fail. A non-zero decimal portion guarantees us that more than one bit is set.

The original poster asked for an example using Siemens, however all I've got handy at the moment is an AB implementation:

-Trevor

Peter Nachtwey · Aug 6, 2010

We have been over this many times.
You guys are inventing the wheel yet again and it isn't very round.
Bernie is close and knows where to go
Did you know you can simply search for "count two bits" on Google?

tvey · Aug 6, 2010

Slightly OT

I don't see a problem answering a question like this simply because the answer can also be found elsewhere.

Sure, 'bit-twiddling' is the preferred technique. When I'm working in C, bits are twiddled all over the place!

There are, however, situations where using an alternate technique may be warranted.

Not all PLC platforms are created equal, and not everyone has the knowledge or interest to implement the purist's solution. In many cases the larger process is of greater concern. A viable, working black-box-ish solution for the specific platform in use could be just what the doctor ordered.

In this case, I invested perhaps 5 minutes into my post, and it provided a much-needed morning break. It's a viable, working solution, as are some of the other suggestions above.

The next guy who searches this forum (or google, for that matter) now has an additional source of information that just might help.

Also - lately I've been getting a kick out of 'Let Me Google That For You': Count Two Bits

Cheers,
Trevor

bernie_carlton · Aug 6, 2010

Actually - if you read the original post - it's not 'TWO BITS' - it's 'MORE THAN ONE'. Which means, after eliminating the first if any are left there are 'MORE THAN ONE'. It doesn't matter if there are exactly two.

TConnolly · Aug 6, 2010

A simple bit hack should do it.

J := I AND NOT(I - 1)
If J<>I then more than one bit is set in I.

----------------------------------------------------------
Example 1

I = 1010

J := I-1 which is 1001
bitwise NOT J to get 0110
bitwise AND J with I to get 0010

J <> I so I <> n² and therefore more than one bit must set.

---------------------------------------------------------------------------
Example 2

I = 0100
J := I-1 which is 0011
bitwise NOT J to get 1100
bitwise AND J with I to get 0100

J = I so I has only one bit set.

TConnolly · Aug 6, 2010

Certain bit hacks are so useful that they are worth memorizing. This is one of them. The j := i AND NOT (i-1) hack has several other uses besides just checking for more than one set bit. I use it in state machines all the time.

Peter Nachtwey · Aug 6, 2010

Getting picky there.

bernie_carlton said:
Actually - if you read the original post - it's not 'TWO BITS' - it's 'MORE THAN ONE'. Which means, after eliminating the first if any are left there are 'MORE THAN ONE'. It doesn't matter if there are exactly two.

Exactly, so stop after finding two bits.

Dectect morethan 1 bit set in a byte.

Darbs003

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erdemsvri

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L D[AR2P#0.0]

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Darbs003

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KalleOlsen

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L D[AR2P#0.0]

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bernie_carlton

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TurpoUrpo

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tvey

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Peter Nachtwey

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tvey

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bernie_carlton

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TConnolly

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