How to reverse an order of an 8 bit value

[esaravanan85]

Hi,

   I want to produce an mirror value of an 8 bit numeric. suppose my value is 11101100, my output should be 00110111. how can i get this??

[JB · Accepted solution]

Here is a solution.

[D_A_N_A_S]

Use Numeric->Data Manipulation -> Rotate

[Yamaeda]

I'd solve it through a small for-loop, bitshift the incoming in one direction (e.g. left) through a shift register and build an outvalue by shifting in values (the output from the former shift) in the other direction (e.g. right).

/Y

[JB]

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Sanad MM a écrit:
Use Numeric->Data Manipulation -> Rotate

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No, this will not produce the expected mirror !

[ST5]

My vi works. It has minor changes to the one posted above.

They're explained in the front panel.

//edit: version for LV8.6 added

Message Edited by ST5 on 03-23-2010 06:54 AM

[esaravanan85]

Thanks JB 🙂 i got my solution....

[altenbach]

What we've learned from the bit twiddling challenge, doing the detour over the boolean array is not very efficient. If this is used on huge amounts of data in tight loops, it might be better to use a lookup table instead. Look ma, no green!!! 🙂

(Of course you would create the lookup table once using a FOR loop with N=256, and containing the above code.)

 

Message Edited by altenbach on 03-24-2010 12:19 AM

[Yamaeda]

I still find my little loop suggestion fun. 😉

/Y

[altenbach]

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Yamaeda wrote:

I still find my little loop suggestion fun. 😉

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Very cute. That's probably the fastest you can do with explicit code.

Here are some benchmarks (5M U8 numbers, LabVIEW 8.0, debugging disabled).

Green (reverse boolean array): 385ms 10x slower

Yamaeda:                       272ms  7x slower

My Lookup table:                39ms     winner.

Clearly, the LUT wins hands down. 😉