Why when I use multiplication
float a= 1.0500000f;
float b= a*100.0f;
why b is 104.99999 but not 105.0 ?
and when I
int f= (int)b;
f is 104
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mrgloom
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3[Is floating point math broken?](http://stackoverflow.com/questions/588004/is-floating-point-math-broken) – 001 Apr 10 '14 at 13:32
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4See [here](http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html) for everything you need to know. As for the senond part, it's due to truncation. – Phylogenesis Apr 10 '14 at 13:32
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there are lots of questions about this "problem" on this site. Read more http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html – phuclv Apr 10 '14 at 13:42
1 Answers
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Floating point numbers are not infinitely accurate -- the wikipedia page on Float numbers is quite interesting to read.
On your second question: (int)b truncates anything that occurs after the comma. This means that, in your case, 104.99999 becomes 104. So when you create a rounding error, and then cast it to an integer, you are indeed running the risk of getting a lower number.
Lee White
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2Somewhat wrong on the second part, casting to `int` doesn't *round* down, it simply *truncates* the value. – Some programmer dude Apr 10 '14 at 13:34
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... the result of which is that the number is rounded down. But yes, I have updated my post. – Lee White Apr 10 '14 at 13:34
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