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Home/ Questions/Q 9127373
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T07:16:04+00:00 2026-06-17T07:16:04+00:00

the code below outputs 0.0. is this because of the overflow? how to avoid

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the code below outputs 0.0. is this because of the overflow? how to avoid it? if not, why?

p ((1..100000).map {rand}).reduce :*

I was hoping to speed up this code:

p r.reduce(0) {|m, v| m + (Math.log10 v)}

and use this instead:

p Math.log10 (r.reduce :*)

but apparently this is not always possible…

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  1. Editorial Team
    Editorial Team
    2026-06-17T07:16:05+00:00Added an answer on June 17, 2026 at 7:16 am

    The values produced by rand are all between 0.0 and 1.0. This means that on each multiplication, your number gets smaller. So by the time you have multiplied 1000 of them, it is probably indistinguishable from 0.

    At some point, ruby will take your number to be so small that it is 0. for instance: 2.0e-1000 # => 0

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