红宝石无符号右移 [英] Ruby Unsigned Right Shift

问题描述

我的java code的一个片段：
收益率（INT）（种子＆GT;＆GT;＆GT;（48 - 位））;

I have a snippet of java code: return (int)(seed >>> (48 - bits));

正如你可以看到它使用了无符号向右移位运算符（>>>）。我想在Ruby中实现这个code有没有符号向右移位运算符，只有签署右移位运算。由于我不是很熟悉>>>运营商我不是我怎么会impelemnt这红宝石十分肯定。我试着做一些搜索，看看有没有人遇到过这个问题，但找不到任何培训相关。任何帮助将是非常美联社preciated：）

As you can see it uses the unsigned right shift operator (>>>). I'm trying to implement this code in ruby which has no unsigned right shift operator, only a signed right shift operator. As I'm not very familiar with the >>> operator I'm not quite sure on how I'd impelemnt this in ruby. I tried doing a few searches to see if anyone had come across this problem before, but couldn't find anything relevent. Any help would be much appreciated :)

推荐答案

如果你想实现一个无符号移位运算，我们可以用一些工作做到这一点：没有，如果有任何我只是觉得自己的简单的解决方案的想法：

If you want to implement an unsigned shift operator we can do this with a bit of work: No idea if there's any simpler solution I just thought of that myself:

public static int unsignedShift(int amt, int val) {
    int lastBit = val & (1 << 31);  // get last bit
    val = val & 0x7fffffff; // clear last bit, ~(1 << 31) could be more efficient
    val >>= amt;   // now we can safely shift signed.
    if (lastBit != 0) {
        val |= 1 << (32 - amt - 1); // set last bit again.
    }
    return val;
}

会好得多，如果是这样！lastBit会工作 - 那么，如果我们就不需要在

Would be much nicer if something like !!lastBit would work - then we wouldn't need the if.

没有关于Ruby的想法，所以你必须端口上面code你自己 - 应该足够简单

No idea about Ruby so you'll have to port the above code yourself - should be simple enough.

编辑：只是想到了一个更简单的方法来做到这一点：（不为0和32正常工作将由JLS的定义 - 也许可以解决的，但最常见如果出现这种情况，无论如何一个bug）

Just thought of a much easier way to do this: (does not work correctly for 0 and 32 which would be defined by the JLS - probably fixable but most often a bug if that happens anyhow)

public static int unsignedShift2(int amt, int val) {
    int mask = (1 << (32 - amt)) - 1;   
    return (val >> amt) & mask;
}

掩模作品通过设置所有位为1应后移被保留（准确地说1转向，应该是0之后，然后通过1减小第一位）。

The mask works by setting all bits to 1 that should be retained after the shift (to be exact shift a 1 to the first bit that should be 0 afterwards and then decrease by 1).

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