Question

Possible Duplicate:
java for-loop problem

Why is the output of following code:

for (float j2 = 0.0f; j2 < 10.0f; j2+=0.1f) {
    System.out.println(j2);
}

this:

0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.70000005
0.8000001
0.9000001
1.0000001
1.1000001
1.2000002
1.3000002
1.4000002
1.5000002
1.6000003
1.7000003
1.8000003
1.9000003
2.0000002
2.1000001
2.2
2.3
2.3999999
2.4999998
2.5999997
2.6999996
2.7999995
2.8999994
2.9999993
3.0999992
3.199999
3.299999
3.399999
3.4999988
3.5999987
3.6999986
3.7999985
3.8999984
3.9999983
4.0999985
4.1999984
4.2999983
4.399998
4.499998
4.599998
4.699998
4.799998
4.8999977
4.9999976
5.0999975
5.1999974
5.2999973
5.399997
5.499997
5.599997
5.699997
5.799997
5.8999968
5.9999967
6.0999966
6.1999965
6.2999964
6.3999963
6.499996
6.599996
6.699996
6.799996
6.899996
6.9999957
7.0999956
7.1999955
7.2999954
7.3999953
7.499995
7.599995
7.699995
7.799995
7.899995
7.9999948
8.099995
8.199995
8.299995
8.399996
8.499996
8.599997
8.699997
8.799997
8.899998
8.999998
9.099998
9.199999
9.299999
9.4
9.5
9.6
9.700001
9.800001
9.900002

Also one more question: even if I change condition of for loop to j2<=10.0f the output is the same. Why so? Shouldn’t it include 10.0 in output?

Answer 1

As others have mentioned it is about the precision of the floating point numbers and the fact that each step of manipulation could potentially increase the error.

This is why you never use floats to represent money. Check out BigDecimal for exact decimal handling.

The reason that 10.0 is not included is that the sum has floated and no longer is <= 10.0 since the precision/rounding error has drifted above +0.0. 9.900002+0.1 will probably yield something like 10.000002 which is > 10.0.

Floats are not represented as exact numbers but rather as a coefficient and an exponent such as (−1)^s × c × b^q and adding to floats will combine uncertainty from both numbers. This might translate to 0.70000005 instead of 0.7 due to loss of precision.

Check out http://en.wikipedia.org/wiki/IEEE_754-2008