AB copy bits from one word to another

[hannibalevivo]

Hi, I'm starting with the ladder logic programming and I want to enssure if the following bitwise operation is correct or not.

I want to copy B9:12/12-9 bits to B3:11/14-11

Here is the sequence of instructions to do this:

CLR B3:11
MVM Source: B9:12 Mask: 1E00h Destination: B3:11
MUL Source A: B3:11 source B: 4 Destination: B3:11
(left shift, 2 positions)

Thanks in advance

 

[widelto]

Hi, I'm starting with the ladder logic programming and I want to enssure if the following bitwise operation is correct or not.

I want to copy B9:12/12-9 bits to B3:11/14-11

Here is the sequence of instructions to do this:

CLR B3:11
MVM Source: B9:12 Mask: 1E00h Destination: B3:11
MUL Source A: B3:11 source B: 4 Destination: B3:11
(left shift, 2 positions)

Thanks in advance

Hi hanni:
Since you´re only copying four bits 12-9 to 14-11 why don´t you copy one by one it´s going to take only four rungs.
Example:
xic [B9:12/12]-------- OTE (B3:11/14)
xic [B9:12/11]-------- OTE (B3:11/13)
xic [B9:12/10]-------- OTE (B3:11/12)
xic [B9:12/09]-------- OTE (B3:11/11)

 

[Ron Beaufort]

I'm starting with the ladder logic programming and I want to ensure if the following bitwise operation is correct or not.

if we assume (gosh I hate that word) that all of the OTHER bits in word B3:11 are to be cleared to a ZERO status, then the code that you posted appears to do what you said you wanted ...



but ... since you’re just starting out, you might also be interested in going one step deeper - and try moving the same four bits into B3:11/15-12 ... in that case, the MUL that you used to “shift” the bit pattern is going to prove “onerous” - and the simple rung-by-rung approach suggested by my distinguished colleague widelto is likely to prove much more reliable ...



in any case, the rung-by-rung approach is probably going to be a lot easier for a maintenance technician to understand whenever a troubleshooting problem eventually pops up ...



good luck on your studies ...