＆是什么QUOT;＆GT;＆GT;＆GT;＆QUOT;在java中是什么意思？ [英] what does ">>>" mean in java?

问题描述

我发现这个code找到<一个副本href=\"http://stackoverflow.com/questions/12372385/binary-search-olog-n-algorithm-to-find-duplicate-in-sequential-list?rq=1\">SO张贴在这里。
但我不明白这是什么意思行 INT中旬=（低+高）GT;＆GT;＆GT; 1;

 私有静态诠释findDuplicate（INT []数组）{
        INT低= 0;
        INT高= array.length  -  1;        而（低＆LT; =高）{
            INT中旬=（低+高）GT;＆GT;＆GT; 1;
            的System.out.println（中）;
            INT midVal =阵列[MID];            如果（midVal ==中旬）
                低=中旬+ 1;
            其他
                高=中旬 -  1;
        }
        高回报;
    }
 

解决方案

的＆GT;＆GT;＆GT; 运算符是的无符号的右位移位在Java中运营商。它有效地把通过 2 操作数右操作数的电源，或者只是 2 在这里。

之间的差异＆GT;＆GT; 和＆GT;＆GT;＆GT; 转向负时，只会出现数字。在＆GT;＆GT; 运营商移动一个 1 位进入最显著位，如果它是一个 1 和＆GT;＆GT;＆gt;在<$ C $ 班C> 0 不管。

更新：

让平均 1 和 2147483647 （ Integer.MAX_VALUE的）。我们可以很容易地算一算：

 （1 + 2147483647）/ 2 =2分之2147483648= 1073741824
 

现在，用code （低+高）/ 2 ，这些都是涉及到的位：

  1：00000000 00000000 00000000 00000001
[phone]：01111111 11111111 11111111 11111111
================================================
-2147483648：千万00000000 00000000 00000000 //溢出
/ 2
================================================
-1073741824：11000000 00000000 00000000 00000000 //符号除法，同＆GT;＆GT; 1。
 

让我们把转移到＆GT;＆GT;＆GT; ：

  1：00000000 00000000 00000000 00000001
[phone]：01111111 11111111 11111111 11111111
================================================
-2147483648：千万00000000 00000000 00000000 //溢出
＆GT;＆GT;＆GT; 1
================================================
1073741824：01000000 00000000 00000000 00000000 //无符号右移。
 

I found this code to find duplicates in SO post here. but I dont understand what this line means int mid = (low + high) >>> 1;

private static int findDuplicate(int[] array) {
        int low = 0;
        int high = array.length - 1;

        while (low <= high) {
            int mid = (low + high) >>> 1;
            System.out.println(mid);
            int midVal = array[mid];

            if (midVal == mid)
                low = mid + 1;
            else
                high = mid - 1;
        }
        return high;
    }

解决方案

The >>> operator is the unsigned right bit-shift operator in Java. It effectively divides the operand by 2 to the power of the right operand, or just 2 here.

The difference between >> and >>> would only show up when shifting negative numbers. The >> operator shifts a 1 bit into the most significant bit if it was a 1, and the >>> shifts in a 0 regardless.

UPDATE:

Let's average 1 and 2147483647 (Integer.MAX_VALUE). We can do the math easily:

(1 + 2147483647) / 2 = 2147483648 / 2 = 1073741824

Now, with the code (low + high) / 2, these are the bits involved:

          1: 00000000 00000000 00000000 00000001
[phone]: 01111111 11111111 11111111 11111111
================================================
-2147483648: 10000000 00000000 00000000 00000000  // Overflow
/2
================================================
-1073741824: 11000000 00000000 00000000 00000000  // Signed divide, same as >> 1.

Let's make the "shift" to >>>:

          1: 00000000 00000000 00000000 00000001
[phone]: 01111111 11111111 11111111 11111111
================================================
-2147483648: 10000000 00000000 00000000 00000000  // Overflow
>>> 1
================================================
[phone]: 01000000 00000000 00000000 00000000  // Unsigned shift right.

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