Java - infinite while loop with float

Question

I am currently learnig java float-point numbers. I know, that a float have a specific amount of significant digets. I also know, that a float is represented in java like -1 or 1 * num * 10^x. Where num is the numer and 10^x is the decimal "point". But here we do not have a fraction of a number. How is a infinite loop here possible?

The code with infinite loop:

float f = 123456789;
while (f-- > 0) {
System.out.println(f);
}

Are you sure the loop is infinite? What happens after f becomes 0? — Thorsten Dittmar, Jan 01 '17 at 10:12
f stays the same and no decrement. It is an infinite loop. It was a exam question at the university last year (example exam). — Charlie, Jan 01 '17 at 10:13
f is not precise enough to represent this number (f--) is the solution but why? — Charlie, Jan 01 '17 at 10:15
Check this http://stackoverflow.com/questions/31903174/why-there-is-loss-of-value-when-converting-from-int-to-float-in-the-below-code — Antoniossss, Jan 01 '17 at 10:35

Adrian Jałoszewski · Accepted Answer · 2017-01-01T10:34:22.343

4

It's about floating point arithmetic. When you're decrementing the number you stumble upon the situation when you go through the number f = 1.23456792E8 makes everything go wrong, because f-1 and f have the same floating point representation here. So decrementing makes the number be itself, which leads to the infinite loop. Just check:

System.out.println(1.23456792E8f);
System.out.println(1.23456792E8f - 1);

What is so special about the number and why execution stops there? It's because the floating point representation of 123456789 is 1.23456792E8. Already this gives us a gap of at least 3 due to the lack of precision of floating point numbers. Using double instead of float makes the program finish, but the issue will occur with higher numbers.

edited Jan 01 '17 at 10:34

answered Jan 01 '17 at 10:25

Adrian Jałoszewski

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what do You mean by destroys everything? – Tahir Hussain Mir Jan 01 '17 at 10:26
the algorithm - thanks, I'll change it – Adrian Jałoszewski Jan 01 '17 at 10:26
why it isn't 0.23456792E8? – Charlie Jan 01 '17 at 10:27
If you mean 1.23456789E8 - the number just has no floating point representation as far as I see, so it has to be the nearest to it. I'll add this to the answer. – Adrian Jałoszewski Jan 01 '17 at 10:31
ahh thank you. But the f-- will decrement something but not in the relevant area? – Charlie Jan 01 '17 at 10:39
`f--` is just short (at least this case) for `f=f−1`, but in this case it will still be `f` because of the lack of precission - it will perform the operation, but it will still be the same number. – Adrian Jałoszewski Jan 01 '17 at 10:56

Java - infinite while loop with float

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