How do you get the mantissa of a float in java?

Question

I'm trying to get the mantissa of a float (just to learn), but it isn't working as expected.

structure of a floating point number

The mantissa of say 5.3 is 53, right? I tried this code:

System.out.println(Float.floatToIntBits(5.3f) & 0x7FFFFF);

It printed 2726298. Shouldn't it remove the exponent bits and leave 53? I tried plenty of things, but this always happens. What am I doing wrong here?

The base of the exponent for an IEEE floating point is *2*, not 10. — awksp, Jun 17 '14 at 22:01
"The mantissa of say 5.3 is 53, right?". Wrong. It's binary, not base 10 — The Archetypal Paul, Jun 17 '14 at 22:02
You're thinking of the mantissa in terms of a *decimal* point - it's a *binary* point... — Jon Skeet, Jun 17 '14 at 22:02
`5.3f` isn't `5.3 x 10^1` in floating point. It's `0b01010011001100110011010 x 2^10000001` (don't know if those are the exact values, but I think what I have is somewhere in the right ballpark) — awksp, Jun 17 '14 at 22:04
@Paul I don't understand, wouldn't there be a binary number for the mantissa that equals 53? — user3452725, Jun 17 '14 at 22:04
`5.2 ~ 2726298 * 2^-19` So it looks you also typoed `5.3` instead of `5.2`. — Mysticial, Jun 17 '14 at 22:07
This whole time I thought it was the exponent for a base 10 number :( — user3452725, Jun 17 '14 at 22:07
That's the whole reason it causes so much confusion -- that it's base 2. That's why for instance you can't store 0.1 exactly in a float (it's an infinitely repeating fraction in binary). — yshavit, Jun 17 '14 at 22:44

CMPS · Accepted Answer · 2014-06-17T22:28:32.443

15

The formula for single precision following the IEEE standard is:

(-1)^sign + 1.Mantissa x 2^(Exponent - Bias)

So 5.3 base 10 is 101.0100110011001100110011 base 2

101.0100110011001100110011 = 1.010100110011001100110011 * 2^2

2^2 = 2^(exp - bias) having bias = 127 (according to the IEEE standard for single precision)
so: exp - 127 = 2 => exp = 129 base 10 or 10000001 base 2

Single precision table:

0 | 10000001 | 01010011001100110011001

Sign = 0
Exp = 129
Mantissa = 2726297

edited Jun 17 '14 at 22:28

answered Jun 17 '14 at 22:09

CMPS

7,733
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Hmmm... Seems I'm interpreting something wrong then. `Integer.toBinaryString(Float.floatToIntBits(5.3f))` prints `1000000101010011001100110011010`, and it doesn't look like the exponent there matches... Where am I going wrong? – awksp Jun 17 '14 at 22:13
How is the bias determined? If it's defined by IEEE and Java is IEEE-compliant shouldn't that not be an issue? – awksp Jun 17 '14 at 22:23
@user3580294 bias is determined by the IEEE standard, for single precision (float) it is 127 and for double precision it is 1023 – CMPS Jun 17 '14 at 22:29
@user3580294 yes, I accidentally flipped the position of Exp and Bias in the formula – CMPS Jun 17 '14 at 22:31

score 1 · Answer 2 · answered Oct 08 '18 at 14:42

1

From the article IBM: Java's new math. Floating-point numbers (in Russian) the simplest way to get the mantissa is:

public static double getMantissa(double x) {
    int exponent = Math.getExponent(x);
    return x / Math.pow(2, exponent);
}

answered Oct 08 '18 at 14:42

Sergey Ponomarev

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How do you get the mantissa of a float in java?

2 Answers2

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