different values for float and double

Question

I don't understand why are float values different from double values. From the example bellow it appears that float provides different result than double for the same operation:

public class Test {

    public static void main(String[] args) {
        double a = 99999.8d;
        double b = 99999.65d;
        System.out.println(a + b);

        float a2 = 99999.8f;
        float b2 = 99999.65f;
        System.out.println(a2 + b2);
    }
}

Output:

199999.45
199999.44

Can you explain what makes this difference between float and double?

http://en.wikipedia.org/wiki/IEEE_754-1985#Single-precision_32_bit — Damian Leszczyński - Vash, Jul 20 '12 at 14:30
What the others said, but in your specific case the result is wider than the number of significant digits possible with floats (cfr wikipedia, 7.2 digits) - see also http://en.wikipedia.org/wiki/Single_precision — fvu, Jul 20 '12 at 14:39

Ross Larson · Accepted Answer · 2012-07-20T15:09:17.453

25

A float is a 32 bit IEEE 754 floating point.

A double is a 64 bit IEEE 754 floating point.

so it is just a matter of precision because neither of the fraction portions .8 and .65 have a terminating binary representation, so there is some rounding error. the double has more precision so it has slightly less rounding error.

http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html

edited Jul 20 '12 at 15:09

answered Jul 20 '12 at 14:28

Ross Larson

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score 14 · Answer 2 · answered Jul 20 '12 at 14:36

Can you explain what makes this difference between float and double?

Sure. Imagine you had two decimal types, one with five significant digits, and one with ten.

What value would you use to represent pi for each of those types? In both cases you'd be trying to get as close to a number which you couldn't represent exactly - but you wouldn't end up with the same value, would you?

It's the same for float and double - both are binary floating point types, but double has more precision than float.

+1 A better example; `one with 6 significant digits and one with 16` ;) — Peter Lawrey, Jul 20 '12 at 14:42

score 12 · Answer 3 · edited Dec 14 '15 at 10:41

12

public class Main {

    public static void main(String[] args) {
        float  a=12.6664287277627762f;
        double b=12.6664287277627762;

        System.out.println(a);
        System.out.println(b);
    }
}

Output:

12.666429
12.666428727762776

float can handle about 7 decimal places. A double can handle about 16 decimal places.

edited Dec 14 '15 at 10:41

Saeed Zarinfam

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answered Feb 02 '13 at 09:42

Thavamani

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score 5 · Answer 4 · answered Jul 20 '12 at 14:27

5

Doubles have twice the precision of floats. Thus they have smaller rounding errors.

A float has (usually) 32 bits, and a double 64 (again usually). Thus floats have rounding errors on more numbers than doubles.

answered Jul 20 '12 at 14:27

jhoyla

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4

+1 in Java float is always 32-bit and double is always 64-bit. – Peter Lawrey Jul 20 '12 at 14:41

Caffeinated · Answer 5 · 2012-07-20T14:43:05.497

1

Floats have less precision than doubles.

It's roughly half as much - 23 bits vs 52 for double(Thanks a lot Mr. Skeet!)!

32-bit for floats, 64-bit for doubles. ...Remember that the word "float" has fewer letters than "double", that's a "memory" trick :)

edited Jul 20 '12 at 14:43

answered Jul 20 '12 at 14:30

Caffeinated

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Not "half" - a `float` only has 23 bits of precision, and `double` has 52. It's not like the mantissa size is proportional to the overall size. – Jon Skeet Jul 20 '12 at 14:37

different values for float and double

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