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Um, since a Java int value of specified by the literal '0xffff' has a 32-bit
value of 0x0000FFFF, how does a bitwise AND with such a value not
zero-out the hi-order byte?
- Fernando
>
> The '& 0xffff' won't place 0's in the higher registers. It will just
> leave them there if they're already present. If, on the other hand,
> there are 1's there, it'll leave those too...
>
> But, as you said, it doesnt really matter :-)
>
> -MG
>
> Arkaitz Bitorika wrote:
> > You are supposing that the Java integers are 16 bit long, and I
> think they
> > actually are 32 bit integers, so the & 0xffff operation will put 0s in
> > the higher 16 bits.
> > I have no idea what it is doing this here, but it certainly
> *is* having an
> > effect.
> >
> > Cheers,
> >
> > Arkaitz.
> >
>
>
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