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JD2 Permalink
October 20, 2010, 18:59

First to be clear, Peter' claim was to clear the right-most 1 bit, not to be confused with clearing the LSB: (e.g. 0xABCD & 0xFFFE).

So in both of your examples, which you use binary numbers ending in 10b, the result is that the right-most 1 bit (e.g. the n-1 position) is indeed cleared resulting in 00b.

So what's the problem? The original number x ending in 10b already had 0 as its LSB before the operation and post operation it's still 0. The only thing that changes was the right most 1 bit which was in the n-1 position.

This is not a counter-example. Rather, you've just given more examples to support the operation.

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