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用ES6语法重写一遍然后改地方反馈一下 #2

@GitaiQAQ

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@GitaiQAQ

新的Hash计算函数

function hashU(x, K) {
    x += "";
    for (var N = [], T = 0; T < K.length; T++) N[T % 4] ^= K.charCodeAt(T);
    var U = ["EC", "OK"],
        V = [];
    V[0] = x >> 24 & 255 ^ U[0].charCodeAt(0);
    V[1] = x >> 16 & 255 ^ U[0].charCodeAt(1);
    V[2] = x >> 8 & 255 ^ U[1].charCodeAt(0);
    V[3] = x & 255 ^ U[1].charCodeAt(1);
    U = [];
    for (T = 0; T < 8; T++) U[T] = T % 2 == 0 ? N[T >> 1] : V[T >> 1];
    N = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F"];
    V = "";
    for (T = 0; T < U.length; T++) {
        V += N[U[T] >> 4 & 15];
        V += N[U[T] & 15]
    }
    return V
}

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