我试图将double转换为它的二进制表示,但使用这个Long.toBinaryString(Double.doubleToRawLongBits(d))
没有帮助,因为我有大数字,Long不能存储它们,即2^900
。
Long.toBinaryString(Double.doubleToRawLongBits(d))
似乎工作得很好。
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
*/
您可能想要处理整个和小数部分:
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();
}
您可以使用BigInteger来保存您的大数字和BigInteger.toString()方法来检索它的二进制表示。
BigInteger bigNum = new BigInteger(sYourNum);
System.out.println( bigNum.toString(2) );
您是否尝试过使用java.math.BigInteger
并使用参数2调用toString(int radix)
?
您可以使用Double.toHexString(d),然后使用for循环和StringBuilder将十六进制字符串转换为二进制字符串。