'Convert Double to Binary representation?

I tried to convert a double to its binary representation, but using this Long.toBinaryString(Double.doubleToRawLongBits(d)) doesn't help, since I have large numbers, that Long can't store them i.e 2^900.

javabinarydoubletype-conversionlong-integer


Solution 1:[1]

Long.toBinaryString(Double.doubleToRawLongBits(d)) appears to work just fine.

System.out.println("0:                0b" + Long.toBinaryString(Double.doubleToRawLongBits(0D)));
System.out.println("1:                0b" + Long.toBinaryString(Double.doubleToRawLongBits(1D)));
System.out.println("2:                0b" + Long.toBinaryString(Double.doubleToRawLongBits(2D)));
System.out.println("2^900:            0b" + Long.toBinaryString(Double.doubleToRawLongBits(Math.pow(2, 900))));
System.out.println("Double.MAX_VALUE: 0b" + Long.toBinaryString(Double.doubleToRawLongBits(Double.MAX_VALUE)));

/*
    prints:
    0:                0b0
    1:                0b11111111110000000000000000000000000000000000000000000000000000
    2:                0b100000000000000000000000000000000000000000000000000000000000000
    2^900:            0b111100000110000000000000000000000000000000000000000000000000000
    Double.MAX_VALUE: 0b111111111101111111111111111111111111111111111111111111111111111
*/

Solution 2:[2]

You may want to process whole and fractional part :

public String toBinary(double d, int precision) {
    long wholePart = (long) d;
    return wholeToBinary(wholePart) + '.' + fractionalToBinary(d - wholePart, precision);
}

private String wholeToBinary(long l) {
    return Long.toBinaryString(l);
}

private String fractionalToBinary(double num, int precision) {
    StringBuilder binary = new StringBuilder();
    while (num > 0 && binary.length() < precision) {
        double r = num * 2;
        if (r >= 1) {
            binary.append(1);
            num = r - 1;
        } else {
            binary.append(0);
            num = r;
        }
    }
    return binary.toString();
}

Solution 3:[3]

You can use a BigInteger to hold your large number and the BigInteger.toString() method to retrieve a binary representation of it. 

BigInteger bigNum = new BigInteger(sYourNum);
System.out.println( bigNum.toString(2) );

Solution 4:[4]

Here is a small snippet which converts the fractional part of the double to binary format:

String convertToBinary(double number) {
    int i=1;
    String num="0.";
    double temp,noofbits=32;
    while (i<=noofbits && number>0) {
        number=number*2;
        temp=Math.floor(number);
        num+=(int)temp;
        number=number-temp;
        i++;
    }

where noofbits gives the bitsize you want the fractional part to be limited to. For the whole number part directly use the Integer.toBinaryString() along with the floor value of the double and append to the fractional binary string.

Solution 5:[5]

Have you tried using java.math.BigInteger and calling toString(int radix) with a parameter of 2?

Solution 6:[6]

Though this question is old, and good answers are present.

I just come up with an idea that you can write the double value to a file/memory region through DataOutputStream and read it as bytes.

Therefore we can use ByteArrayOutputStream to hold the binary representation and fetch the bytes directly to avoid I/O operations on disk. (javadoc, also this post)

The scala version of it looks like:

import java.io._

val baos = new ByteArrayOutputStream(8)
val dos = new DataOutputStream(baos)
dos.writeDouble(1.23)
dos.close()

// this 'Array[Byte]' or 'byte[]' in java holds the correct binary representation (big-endian) of the double. 
val binaryRepresentation = baos.toByteArray() 
baos.close()

Solution 7:[7]

You can use Double.toHexString(d) and then transform the hexadecimal string into a binary one using a for loop and a StringBuilder.

Sources

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Source: Stack Overflow

Solution Source
Solution 1
Solution 2 Bax
Solution 3 Brandon E Taylor
Solution 4 gawi
Solution 5 Liv
Solution 6 Aria Ax
Solution 7 Aymen

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