From: Manu Sankala <manuska@...>
Date: 2010-02-22T03:19:21+09:00
Subject: Re: Yet another rounding issue

Mohit Sindhwani wrote:
> You could try:
> a = 2.175
> b = a + 0.0005
> c = ((b * 100).round) / 100.0

Well this is basically same as my
((some_float_here*100).to_s+'9').to_f.round/100.0
except that in my version I don't need to know how many decimals the 
float originally has. "b = a + 0.0005" would fail if 'a' has mode than 3 
decimals.

Since
((float_here*100).to_s+'9').to_f.round/100.0
and
(BigDecimal(float_here.to_s)*100).round.to_s.to_f/100
seem to be about as fast I think I'll go with bigfloat as it seems less 
like a hack.

Thank you both for your quick answers

-
Manu S
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