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From "Phil Steitz (JIRA)" <j...@apache.org>
Subject [jira] Commented: (LANG-381) NumberUtils.min(floatArray) returns wrong value if floatArray[0] happens to be Float.NaN
Date Sun, 02 Dec 2007 20:43:43 GMT

    [ https://issues.apache.org/jira/browse/LANG-381?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel#action_12547665
] 

Phil Steitz commented on LANG-381:
----------------------------------

The comparator has to impose a total ordering, so it has to put NaN somewhere.  The min/max
functions do not have to do this - i.e., we could define the contract of min/max to be to
return NaN iff there are no non-NaN values in the array, so NaNs are effectively excluded.
 This is what we did in [math].  See, e.g., 
http://commons.apache.org/math/api-1.1/org/apache/commons/math/stat/StatUtils.html#min(double[],%20int,%20int)

> NumberUtils.min(floatArray) returns wrong value if floatArray[0] happens to be Float.NaN
> ----------------------------------------------------------------------------------------
>
>                 Key: LANG-381
>                 URL: https://issues.apache.org/jira/browse/LANG-381
>             Project: Commons Lang
>          Issue Type: Bug
>    Affects Versions: 2.3
>            Reporter: Thomas Vandahl
>             Fix For: 2.4
>
>
> The min() method of NumberUtils returns the wrong result if  the first value of the array
happens to be Float.NaN. The following code snippet shows the behaviour:
>         float a[] = new float[] {(float) 1.2, Float.NaN, (float) 3.7, (float) 27.0, (float)
42.0, Float.NaN};
>         float b[] = new float[] {Float.NaN, (float) 1.2, Float.NaN, (float) 3.7, (float)
27.0, (float) 42.0, Float.NaN};
>         
>         float min = NumberUtils.min(a);
>         System.out.println("min(a): " + min); // output: 1.2
>         min = NumberUtils.min(b);
>         System.out.println("min(b): " + min); // output: NaN
> This problem may exist for double-arrays as well. 
> Proposal: Use Float.compare(float, float) or NumberUtils.compare(float, float) to achieve
a consistent result.

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