Quick Links: Download Gideros Studio | Gideros Documentation | Gideros Development Center | Gideros community chat | DONATE
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'.

Comments

Sign In or Register to comment.