如何高效地找到一个大整数中的一个字节的值

发布于 2024-10-26 07:56:56 字数 367 浏览 1 评论 0原文

我想找到一个大整数中一个字节的值。例如

11973777 = 10110110 10110100 10010001

我想找到的

11973777[2] = 10110110 = 182
11973777[1] = 10110100 = 180 
11973777[0] = 10010001 = 145

数字并且有效地做到这一点（按位运算）。我可以通过按位运算轻松找到数字中任何位置的位值，但我不想仅仅为了获取字节而执行 8 次，或者先左移然后右移。这被标记为 javascript，因为我正在 javascript 中执行此操作，但我认为可以翻译按位运算。预先感谢您提供的任何帮助。

原文

I want to find the value of a byte in a large integer. For example the number

11973777 = 10110110 10110100 10010001

I would like to find that

11973777[2] = 10110110 = 182
11973777[1] = 10110100 = 180 
11973777[0] = 10010001 = 145

And do so efficiently(bitwise operations). I can easily find the value of a bit at any position in the number with a bitwise operation but I don't want to do that 8 times just to get the byte, or shift left then shift right. This is tagged as javascript because I am doing it in javascript but I suppose bitwise operations can be translated. Thank you in advance for any help you can lend.

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漆黑的白昼 2024-11-02 07:56:56

在某种程度上，你必须进行位移：

var num = 11973777;
var lastByte = num & 0xFF;
var midLowByte = (num >> 8) & 0xFF;
var midHighByte = (num >> 16) & 0xFF;
var highByte = (num >> 24) & 0xFF;

但你总是可以将逻辑封装在函数中以使其更方便，例如：

window.getBytes = function(num) {
    return [num & 0xFF, (num >> 8) & 0xFF, (num >> 16) & 0xFF, (num >> 24) & 0xFF];
};

var numBytes = getBytes(11973777);
alert(numBytes[0] + "," + numBytes[1] + "," + numBytes[2] + "," + numBytes[3]);

To a certain extent, you have to bit-shift:

var num = 11973777;
var lastByte = num & 0xFF;
var midLowByte = (num >> 8) & 0xFF;
var midHighByte = (num >> 16) & 0xFF;
var highByte = (num >> 24) & 0xFF;

But you can always package the logic in a function to make it more convenient, like:

window.getBytes = function(num) {
    return [num & 0xFF, (num >> 8) & 0xFF, (num >> 16) & 0xFF, (num >> 24) & 0xFF];
};

var numBytes = getBytes(11973777);
alert(numBytes[0] + "," + numBytes[1] + "," + numBytes[2] + "," + numBytes[3]);

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