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Question by majakthecoder · Nov 06, 2013 at 12:53 PM · floatprecision

float precision, wrong computations

when I run following code:

 float v = 100.0f - 99.2f;

I get v = 0,8000031 instead of 0.8.

how to increase float precision?

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avatar image ozturkcompany · Nov 06, 2013 at 01:05 PM 1
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That actually pretty precise

avatar image tanoshimi · Nov 06, 2013 at 01:07 PM 1
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That's the inevitable imprecise nature of floating-point arithmetic. If you say what you want the value v to represent, we might be able to suggest alternatives - does that 0.000031 error really matter?

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Answer by Azrapse · Nov 06, 2013 at 01:05 PM

That is what the double type is for.

Although it will use more memory, and also it will keep on having small imprecisions, although much smaller ones. (For example, 0.800000000000000000000000000023).

Also, check this link: link text

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avatar image majakthecoder · Nov 06, 2013 at 01:09 PM 0
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negative, double also is not equal to 0.8

avatar image Dave-Carlile · Nov 06, 2013 at 01:21 PM 0
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What are you trying to accomplish? If you're trying to compare a floating point result to something, it's rare that you can do it by comparing for equality. Either use @Azrapse's suggestion about using $$anonymous$$athf.Approximately, or use your own to deal with small (nearly always meaningless) differences, e.g. ins$$anonymous$$d of comparing for equality with 0, do something like this...

 float epsilon = 0.0001f;

 if (value <= epsilon)
 {
   // consider the value to be 0 and do something
 }

 


avatar image Dave-Carlile · Nov 06, 2013 at 01:25 PM 0
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Also, here is a great discussion on this issue and the very valid reasons why it is this way: http://floating-point-gui.de/

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