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I need to compare extremely small numbers for an artificial intelligence project. Here is my method:

private static int maximum(BigDecimal a1, BigDecimal a2){
    System.out.println(a1);
    System.out.println(a2);
    if (a1.compareTo(a2)<0){
        return 1;
    }else{
         return 0;
    }

Here is my output with the return value printed 3rd:

0E-917
0E-912
0
0E-918
0E-921
0
0E-932
0E-933
0

Is there any way to compare numbers this small or some simple mistake I'm making with the compareTo method? Thanks for the help!

Keagan Galbraith
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1 Answers1

6

0Ewhatever is still 0. They're all zero.

The reason that E-932 even survives at all is because BigDecimal doesn't store just the number, it also knows what precision you're at; 0E-917 is 0, but at a certain precision.

As the spec states, compareTo checks the actual value and disregards the precision level, thus, all of these values are 100% equal to each other: They are all 0.

EDIT: To be clear, 0E-917 is not "an extremely small number". It is zero. That precision thing is effectively: Some calculation was 0.00000000....000000, with 917 zeroes. It might be non-zero, but if it is, it has a non-zero digit only after at least 917 zeroes, and I didn't calculate that far. This, your BD instance has absolutely no way to know that (it just stored 918 zeroes).

rzwitserloot
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