Re^2: Why the Integer part of this calculation is 102?

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Re^3: Why the Integer part of this calculation is 102? by almut (Canon) on Mar 22, 2010 at 21:58 UTC
A bit lengthy and mathematical, but in case you really want to know...: What Every Computer Scientist Should Know About Floating-Point Arithmetic (as PDF).	[reply]
Re^3: Why the Integer part of this calculation is 102? by ikegami (Patriarch) on Mar 22, 2010 at 21:56 UTC
Both 3408/10000 and 9936/10000 are periodic numbers in binary, just like 1/3 is a periodic number in decimal. It would take infinite storage to store them as floats, so they're not being stored exactly. You can avoid the problem by using `sprintf("%.0f", $x)` (rounds) instead of `int($x)` (truncates).	[reply] [d/l] [select]
Re^3: Why the Integer part of this calculation is 102? by moritz (Cardinal) on Mar 22, 2010 at 22:18 UTC
Others have answered you already; I just want to add that in Perl 6 your code does what you expect. It achieves that by storing decimal numbers as fractions/rationals internally. `$ perl6 > say (42435.3408).perl 26522088/625 > say (42435.3408).WHAT Rat() > say 42435.3408 / 411.9936 103 > printf "%.20f\n", 42435.3408 / 411.9936 103.00000000000000000000` [download] Perl 6 - links to (nearly) everything that is Perl 6.	[reply] [d/l]
Re^4: Why the Integer part of this calculation is 102? by alokam (Initiate) on Mar 23, 2010 at 19:13 UTC
Thank you all for your quick response on my issue. Greatly appreciate your help. -Anju	[reply]