What's wrong with using == to compare floats in Java?

the correct way to test floats for 'equality' is:

if(Math.abs(sectionID - currentSectionID) < epsilon)

where epsilon is a very small number like 0.00000001, depending on the desired precision.

Floating point values can be off by a little bit, so they may not report as exactly equal. For example, setting a float to "6.1" and then printing it out again, you may get a reported value of something like "6.099999904632568359375". This is fundamental to the way floats work; therefore, you don't want to compare them using equality, but rather comparison within a range, that is, if the diff of the float to the number you want to compare it to is less than a certain absolute value.

This article on the Register gives a good overview of why this is the case; useful and interesting reading.

Just to give the reason behind what everyone else is saying.

The binary representation of a float is kind of annoying.

In binary, most programmers know the correlation between 1b=1d, 10b=2d, 100b=4d, 1000b=8d

Well it works the other way too.

.1b=.5d, .01b=.25d, .001b=.125, ...

The problem is that there is no exact way to represent most decimal numbers like .1, .2, .3, etc. All you can do is approximate in binary. The system does a little fudge-rounding when the numbers print so that it displays .1 instead of .10000000000001 or .999999999999 (which are probably just as close to the stored representation as .1 is)

Edit from comment: The reason this is a problem is our expectations. We fully expect 2/3 to be fudged at some point when we convert it to decimal, either .7 or .67 or .666667.. But we don't automatically expect .1 to be rounded in the same way as 2/3--and that's exactly what's happening.

By the way, if you are curious the number it stores internally is a pure binary representation using a binary "Scientific Notation". So if you told it to store the decimal number 10.75d, it would store 1010b for the 10, and .11b for the decimal. So it would store .101011 then it saves a few bits at the end to say: Move the decimal point four places right.

(Although technically it's no longer a decimal point, it's now a binary point, but that terminology wouldn't have made things more understandable for most people who would find this answer of any use.)