问题
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.
回答1:
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();
}
回答2:
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
回答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.
回答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().
回答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.
来源:https://stackoverflow.com/questions/756430/how-do-i-convert-a-decimal-fraction-to-binary-in-java