Calculate power for int64 exponent precisely - #289
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hcirellu
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March 11, 2026 11:49
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Related: #362, which basically proposes using the canonical |
MichaelChirico
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I don't find quite as big an improvement as in the original finding, but we're about par with it: |
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OK, I'm happy enough with this now. We can iterate more if needed. The seed of your idea ( |
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I added a precise calculation for
POW64with an overflow detection.The example from #288 now returns the correct values.
The results for double exponent are unchanged, i.e.:
In addition, the new implementation is faster, too.
Closes #288