If you go with a representation-independent 'bitlen', I would suggest returning NaN for any negative number.
This is consistent with a simple mathematical definition: bitlen(n) := ceil(log_2(n + 1)) (Or, in js: Math.ceil(Math.log(n + 1) / Math.LN2), which potentially has rounding errors, but works for integers up to 2^48.) Second best preference for me would be to return bitlen(Math.abs(n)). None of this applies if you opt for a representation-dependent (clz-style) bitlen, Nick On 17 January 2014 03:20, Brendan Eich <[email protected]> wrote: > Jason Orendorff wrote: > >> > What is Math.bitlen(-1) then? Isn’t this just the same problem as >>> before, except it happens for negative numbers instead of positive? >>> >> >> No opinion. >> > > Not sure it matters. We could return -1 for any negative number, 0 for 0, > and > 0 for positive integral values. > > /be > > _______________________________________________ > es-discuss mailing list > [email protected] > https://mail.mozilla.org/listinfo/es-discuss >
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