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by Someone
2364 days ago
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This doesn’t solve the general problem that binary floating point cannot exactly represent decimal fractions. The first example new Double('0.3').sub(new Double('0.1')).toNumber()
does the equivalent of float f = (float)(0.3 - 0.1);
which happens to produce a ‘float32’ that prints as “0.2”, but for ‘float64’ vs ‘float128’.I expect that adding a sufficient number of zeroes, as in double f = (float128)0.00003 - (float128)0.00001);
(actual number of zeroes is too low here, to prevent wrapping on phone screens) will surface the problem again. |
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