如何通过最小的增量来改变双倍 [英] How to alter double by its smallest increment

问题描述

是否有坏事或我不明白发生了什么？

  static String getRealBinary（double val）{
 long tmp = Double.doubleToLongBits（val）; 
 StringBuilder sb = new StringBuilder（）; 
 
（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; （int i = 3;  -  i> = 0;）{
 d + = Double.MIN_VALUE; 
 
 System.out.println（d + getRealBinary（d））; 
} 
} 
} 
  

带有输出：

  2.0 [1] [00000000000] [000000000000000000000000000000000000000000000000000] 
 2.0 [1] [00000000000] [000000000000000000000000000000000000000000000000000] 
 2.0 [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] 0000000000000000000000000000000000000000000000000000011] 
  

解决方案

它的长表示（使用 doubleToLo ngBits ，如您在 getRealBinary 中所做的那样），将其长度增加1，最后将新的long返回到通过 longBitsToDouble 。

编辑：Java（从1.5开始）提供 Math.ulp（double） code>，我猜你可以用它来直接计算下一个更高的值： x + Math.ulp（x）。

Is something broken or I fail to understand what is happening?

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));
        }
    }
}

With output:

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]

解决方案

The general idea is first convert the double to its long representation (using doubleToLongBits as you have done in getRealBinary), increment that long by 1, and finally convert the new long back to the double it represents via longBitsToDouble.

EDIT: Java (since 1.5) provides Math.ulp(double), which I'm guessing you can use to compute the next higher value directly thus: x + Math.ulp(x).

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