Convert Double to Binary representation?

Question

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 nu

Answer 1

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();
}

Answer 2

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
*/

Answer 3

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

Answer 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.

Answer 5

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) );

Answer 6

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