如何在Java中将小数转换为二进制？

Question

I need to convert 0.5 in base 10 to base 2 (0.1). I have tried using

Double.doubleToRawLongBits(0.5)

and it returns 4602678819172646912 which I guess is in hex, but it does not make sense to me.

Answer 1

No. 4602678819172646912 is in dec, hex is 0x3fe0000000000000. To dismantle that:

   3   |   F   |   E   |  0 ...
0 0 1 1 1 1 1 1 1 1 1 0 0 ...
s|  exponent         |  mantissa

s is the sign bit, exponent is the exponent shifted by 2^9 (hence this exponent means -1), mantissa is the xxx part of the number 1.xxx (1. is implied). Therefore, this number is 1.000...*2^-1, which is 0.5.

Note that this describes the "normal" numbers only, so no zeros, denormals, NaNs or infinities

Answer 2

Multiply you number by 2^n, convert to an BigInteger, convert to binary String, add a decimal point at position n (from right to left).

Example (quick & ++dirty):

private static String convert(double number) {
    int n = 10;  // constant?
    BigDecimal bd = new BigDecimal(number);
    BigDecimal mult = new BigDecimal(2).pow(n);
    bd = bd.multiply(mult);
    BigInteger bi = bd.toBigInteger();
    StringBuilder str = new StringBuilder(bi.toString(2));
    while (str.length() < n+1) {  // +1 for leading zero
        str.insert(0, "0");
    }
    str.insert(str.length()-n, ".");
    return str.toString();
}

Answer 3

This is decimal for 0x3FE0_0000_0000_0000 . The mantissa is the list of zeros after 3FE (which codes sign and exponent). This is what you are looking for, given that 0.1 before the zeros is implicit.

Answer 4

Do you want to convert the decimal string to floating-point binary or to a binary string? If the former, just use valueOf(); if the latter, use valueOf() followed by toString() or printf().

Answer 5

0.1 is NOT a binary representation of 0.5

Java will represent 0.5 using IEEE 754, as specified on the Java Language Specification . BigInteger.valueOf(Double.doubleToRawLongBits(0.5)).toByteArray() will give you a byte per byte representation of 0.5 as Java does internally.