用64位整数中的JavaScript中的位置XOR

发布于 2025-02-07 09:30:40 字数 1048 浏览 4 评论 0 原文

我正在寻找一种在JavaScript中64位整数上执行XOR的方法。

JavaScript将将其所有双重值投入到签名的32位整数中，以执行位置操作。

function and(v1, v2) {
    var hi = 0x80000000;
    var low = 0x7fffffff;
    var hi1 = ~~(v1 / hi);
    var hi2 = ~~(v2 / hi);
    var low1 = v1 & low;
    var low2 = v2 & low;
    var h = hi1 & hi2;
    var l = low1 & low2;
    return h*hi + l;
}

对于Xor func来说，这是我做过的转换，但无法正常工作，任何帮助修改的帮助都会对我有所帮助。

function XOR(v1, v2) {
    var hi = 0x80000000;
    var low = 0x7fffffff;
    var hi1 = ~~(v1 / hi);
    var hi2 = ~~(v2 / hi);
    var low1 = v1 ^ low;
    var low2 = v2 ^ low;
    var h = hi1 ^ hi2;
    var l = low1 ^ low2;
    return h * hi - l;
}

任何帮助都将受到赞赏。

原文

I am looking for a way of performing a bitwise XOR on a 64 bit integer in JavaScript.

JavaScript will cast all of its double values into signed 32-bit integers to do the bitwise operations.

I already found this function in bitwise AND in Javascript with a 64 bit integer
for bitwise AND and seems to work ok, but still need the XOR.

function and(v1, v2) {
    var hi = 0x80000000;
    var low = 0x7fffffff;
    var hi1 = ~~(v1 / hi);
    var hi2 = ~~(v2 / hi);
    var low1 = v1 & low;
    var low2 = v2 & low;
    var h = hi1 & hi2;
    var l = low1 & low2;
    return h*hi + l;
}

For the XOR Func this is the transformation i have done but is not working properly, any help modifying it would help me.

function XOR(v1, v2) {
    var hi = 0x80000000;
    var low = 0x7fffffff;
    var hi1 = ~~(v1 / hi);
    var hi2 = ~~(v2 / hi);
    var low1 = v1 ^ low;
    var low2 = v2 ^ low;
    var h = hi1 ^ hi2;
    var l = low1 ^ low2;
    return h * hi - l;
}

Any help is appreciated.

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╰つ倒转 2025-02-14 09:30:41

线路上犯了一个错误

var low1 = v1 ^ low;
var low2 = v2 ^ low;

您在应该的，并且＆amp;要恢复数字的下部，然后在两侧执行 ^ XOR运算符

var low1 = v1 & low;
var low2 = v2 & low;

，最后一行应在示例中与＆amp;运营商总和：

return h * hi + l;

上面的完整代码：

function XOR(v1, v2) {
    var hi = 0x80000000;
    var low = 0x7fffffff;
    var hi1 = ~~(v1 / hi);
    var hi2 = ~~(v2 / hi);
    var low1 = v1 & low;
    var low2 = v2 & low;
    var h = hi1 ^ hi2;
    var l = low1 ^ low2;
    return h * hi + l;
}

const res = XOR(114616149816979, 383577466862359);
console.log(res);

You have a mistakes in the lines

var low1 = v1 ^ low;
var low2 = v2 ^ low;

where it should be and & to recover the lower part of the number, and later perform the ^ xor operator in both sides

var low1 = v1 & low;
var low2 = v2 & low;

and the last line should be at the example with & operator a sum:

return h * hi + l;

Complete code above:

function XOR(v1, v2) {
    var hi = 0x80000000;
    var low = 0x7fffffff;
    var hi1 = ~~(v1 / hi);
    var hi2 = ~~(v2 / hi);
    var low1 = v1 & low;
    var low2 = v2 & low;
    var h = hi1 ^ hi2;
    var l = low1 ^ low2;
    return h * hi + l;
}

const res = XOR(114616149816979, 383577466862359);
console.log(res);

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