A Question on "REV"

Howdy

I have not tried this yet, but am confused on the definition of "REV" on page 39 of the PBP Compiler manual, section 4.17.11

" REV reverses the order of the lowest bits in a value. The number of bits to be reversed is from 1 to 32. "

The example shows:

B0 = %10101100 REV 4 'sets B0 to %00000011

I thought only the lowest 4 bits would be reversed - why are bits 5 and 7 also changed? (set to 0) ?

I Suppose I could just write a short program and try it . . . . .

Ken

Yup, that's the way it works.

It reverses the bits by rotating them into a 0'd variable.

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If you want to preserve the upper 4 bits, you can OR them back in.

B0 = (%10101100 & $F0) | (%10101100 REV 4)

--- Same thing with the value in a variable ---
B1 = %10101100
B0 = (B1 & $F0) | (B1 REV 4)

result = 10100011
<br>

Thanks Darrel.

I totally understand - I guess rotate would have made more sense than reverse - anyway thanks

Ken