java - 如何以最小增量改变 double

Question

是什么东西坏了还是我不明白发生了什么？

static String getRealBinary(double val) {
    long tmp = Double.doubleToLongBits(val);
    StringBuilder sb = new StringBuilder();

    for (long n = 64; --n > 0; tmp >>= 1)
        if ((tmp & 1) == 0)
            sb.insert(0, ('0'));
        else
            sb.insert(0, ('1'));

    sb.insert(0, '[').insert(2, "] [").insert(16, "] [").append(']');
    return sb.toString();
}

public static void main(String[] argv) {
    for (int j = 3; --j >= 0;) {
        double d = j;
        for (int i = 3; --i >= 0;) {
            d += Double.MIN_VALUE;
            System.out.println(d +getRealBinary(d));
        }
    }
}

输出:

2.0[1] [00000000000] [000000000000000000000000000000000000000000000000000]
2.0[1] [00000000000] [000000000000000000000000000000000000000000000000000]
2.0[1] [00000000000] [000000000000000000000000000000000000000000000000000]
1.0[0] [11111111110] [000000000000000000000000000000000000000000000000000]
1.0[0] [11111111110] [000000000000000000000000000000000000000000000000000]
1.0[0] [11111111110] [000000000000000000000000000000000000000000000000000]
4.9E-324[0] [00000000000] [000000000000000000000000000000000000000000000000001]
1.0E-323[0] [00000000000] [000000000000000000000000000000000000000000000000010]
1.5E-323[0] [00000000000] [000000000000000000000000000000000000000000000000011]

Answer 1

一般的想法是首先将 double 转换为它的 long 表示(使用 doubleToLongBits，就像您在 getRealBinary 中所做的那样)，将 long 递增 1，最后转换new long 回到它通过 longBitsToDouble 表示的 double。

编辑:Java(自 1.5 起)提供了 Math.ulp(double)，我猜您可以使用它直接计算下一个更高的值:x + Math.ulp (x).