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How can I zero out the last 32 bits of a double, in java, as efficiently as possible?

StuperUser
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user1214927
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    Why would you want to zero out the last 32 bits of a double? – JAB May 16 '14 at 17:48
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    `(x>>32)<<32` ? – But I'm Not A Wrapper Class May 16 '14 at 17:50
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    You wouldn't want to this on a double since a double is encoded as a coefficient and an exponent. Doing a bitwise operation on a double isn't likely to give you the result you expect. Scaling to a float or converting the number to fixed point might work, depending on the value range. – Baldy May 16 '14 at 17:56
  • You can always do `a = Double.longBitsToDouble(Double.doubleToLongBits(a) & 0xFFFFFFFF00000000L);` – Alexis C. May 16 '14 at 17:58
  • Ignore my last comment...this question should help you convert doubles to bits. That way you can use some bit wise operators, then convert back to a double. http://stackoverflow.com/questions/4211095/xoring-two-doubles-in-java – But I'm Not A Wrapper Class May 16 '14 at 18:00

2 Answers2

7

Try this.

private static final long ZERO_OUT_LAST_32_BITS = 0xffffffff00000000L;

public static void main(String[] args) {
    double number = 2.5;

    long numberBits = Double.doubleToLongBits(number);
    double result = Double.longBitsToDouble(numberBits & ZERO_OUT_LAST_32_BITS);
}

It zeroes out the last 32 bits of the binary representation of your double by using a binary AND operation.

However, make sure you know exactly what and why you are doing - I would at least expect an explaining comment next to code like this.

Petr Janeček
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5

Convert to long and mask away the desired bits, convert back:

long l = Double.doubleToLongBits();
l &= 0xFFFFFFFF00000000L;
double truncated = Double.longBitsToDouble(l);

Although I'm not sure what you're trying to achive by that...

Durandal
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