java - 将 Java double 映射到 String 并保留排序顺序

Question

我正在寻找方法String doubleToString(double d) ，满足以下条件:

对于任意一对 double d1和d2如果Double.compare(d1, d2) < 0然后doubleToString(d1).compareTo(doubleToString(d2)) < 0 .

为了说明需求，我为 int 编写了类似的方法类型:

public static String intToString(int i) {
    String string = Integer.toHexString(i - Integer.MIN_VALUE);
    return "00000000".substring(string.length()) + string;
}

结果(每一个 String 都大于前一个):

intToString(-2147483648) = intToString(0x80000000) = "00000000"
intToString(-10)         = intToString(0xfffffff6) = "7ffffff6"
intToString(-1)          = intToString(0xffffffff) = "7fffffff"
intToString(0)           = intToString(0x0)        = "80000000"
intToString(1)           = intToString(0x1)        = "80000001"
intToString(10)          = intToString(0xa)        = "8000000a"
intToString(2147483647)  = intToString(0x7fffffff) = "ffffffff"

我想要的是 double 具有相同的行为方法的。

Answer 1

您可以将 double 转换为长整型数，然后使用与处理整数类似的方法。我假设以下方法可以按您的意愿工作(NaN 被视为最大数字):

public static String doubleToString(double d) {
    final long bits = Double.doubleToLongBits(d);
    final String s = Long.toString(bits);
    return (bits < 0 ? "--------------------" : "00000000000000000000").substring(s.length()) + s;
}

输入和输出示例:

[-Infinity, -1.7976931348623157E308, -4.9E-324, -0.0, 0.0, 4.9E-324, 1.7976931348623157E308, Infinity]
----4503599627370496
----4503599627370497
-9223372036854775807
-9223372036854775808
00000000000000000000
00000000000000000001
09218868437227405311
09218868437227405312