Confusion in output

Question

double d = 2.2;          
System.out.println( d == 2.2f );gives output as false
and
double d = 2.0;          
System.out.println( d == 2.0f );gives output as true.
Why is the first output false. isn't float converted to double before performing d==2.2f and since float is getting converted to double there shouldnt be any loss of precision. isn't it?

Answer 1

It gets even more confusing:
System.out.println(1.25d == 1.25f);Prints true, even 'though there are places after the decimal point.
The problem is, that those numbers you used (1.1, 1.2, ...) are not representable as binary numbers with a finite number of digits (they are repeating the same way that 1/3 is repeating in the decimal system).
Now let's assume for a moment you've got a number that looks like this in binary:
0.110 (i.e. the "110" part is repeating)If you now print 10 digits of that number it looks like this:
0.1101101101If you print 20 digits of that number it looks like this:
0.11011011011011011011That similar to how float and double work in Java (and pretty much every other language out there). Now let's assume you want to compare those two numbers. First you need to bring them to the same format, which you do by appending zeroes to the end. The result will look like this:
0.11011011010000000000 and
0.11011011011011011011Those two numbers are obviously different, so "==" will return false. The moral of the story is:
Don't use "==" comparison if you are doing any calculation and/or conversion with floating point numbers, always check for a (small) range of values instead.
(Note: I made some important simplifications and probably some erros as well, but that should describe the principles at work here).
Read What Every Computer Scientist Should Know About Floating-Point Arithmetic for a more complete and correct explanation, that's probably a bit harder to read.