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Module:NumberAbbr: Difference between revisions

MisterChuChu (talk | contribs)
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MisterChuChu (talk | contribs)
Fix a bug related to calculating illion 999
 
Line 79: Line 79:
if n>=math.huge then
if n>=math.huge then
return math.huge
return math.huge
elseif n>=999 then
elseif n>999 then
i=10^(3*math.floor(math.log10(n)/3))
i=10^(3*math.floor(math.log10(n)/3))
return (math.floor(n/i)>1 and p.illion({args={math.floor(n/i),false}}) or "")..(math.floor(math.log10(n)/3)<=#e3 and e3[math.floor(math.log10(n)/3)] or "["..math.floor(math.log10(n)/3).."]")..(r<3 and p.illion({args={n%i,nil,true,r+1}}) or "")..(r==3 and "..." or "") --it can only go so far down, this limit is to prevent not enough memory errors
return (math.floor(n/i)>1 and p.illion({args={math.floor(n/i),false}}) or "")..(math.floor(math.log10(n)/3)<=#e3 and e3[math.floor(math.log10(n)/3)] or "["..math.floor(math.log10(n)/3).."]")..(r<3 and p.illion({args={n%i,nil,true,r+1}}) or "")..(r==3 and "..." or "") --it can only go so far down, this limit is to prevent not enough memory errors
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