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a=b//c VS a=math.floor(b/c) — Gideros Forum

a=b//c VS a=math.floor(b/c)

keszeghkeszegh Member
edited August 2022 in Bugs and issues
I've realized that if b/c is e.g. half of a negative integer then the above two do not give the same result. (although perhaps they should? at least for me that would be logical)
the reason is that // rounds 'towards 0' and not 'to the lower value'.


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