无符号右移零位意味着什么?

时间:2018-06-04 18:06:12

标签: javascript c++

我需要将Javascript函数移植到C ++。在移植过程中,我找到了这行代码:

return (randseed[3]>>>0) / ((1 << 31)>>>0);

我的问题是:无符号右移零字节到右边意味着什么?移动零字节有什么意义?那相当于它的C / C ++会是什么?

这是我移植的功能:

    function rand() {
            // based on Java's String.hashCode(), expanded to 4 32bit values
            var t = randseed[0] ^ (randseed[0] << 11);

            randseed[0] = randseed[1];
            randseed[1] = randseed[2];
            randseed[2] = randseed[3];
            randseed[3] = (randseed[3] ^ (randseed[3] >> 19) ^ t ^ (t >> 8));

            return (randseed[3]>>>0) / ((1 << 31)>>>0);
    }

(资料来源:https://github.com/ethereum/blockies/blob/master/blockies.js

这是我目前的端口:

double_t Accounts::rand(int *rand_seed) {
    int v1,v2;
    v1=rand_seed[0];
    v2=rand_seed[0] << 11;
    int t=v1 ^ v2;

    rand_seed[0]=rand_seed[1];
    rand_seed[1]=rand_seed[2];
    rand_seed[2]=rand_seed[3];
    v1=rand_seed[3] >> 19;
    v2=t>>8;
    rand_seed[3]=rand_seed[3] ^ v1 ^ t ^ v2;

    uint v=rand_seed[3];
    uint d=1<<31;

    double_t output=v / d;
    return output;
}

也许有C ++知识的人可能会检查我是否正确地做了这件事?

我还在函数内添加了变量的打印输出,期望右移零字节产生不同的结果

      var before,after
      before=randseed[3]     
      after=(randseed[3]>>>0)
      console.log('before='+before+', after='+after)
      return (randseed[3]>>>0) / ((1 << 31)>>>0);

但是console.log显示没有区别:

before=1458668765, after=1458668765
before=1818856068, after=1818856068
before=622943643, after=622943643

0 个答案:

没有答案