如何在javascript中交换变量的endian-ness（字节顺序） [英] How do I swap endian-ness (byte order) of a variable in javascript

问题描述

我正在接收并发送两个小端数字的十进制表示。我想：

I am receiving and sending a decimal representation of two little endian numbers. I would like to:

• 将一个变量向左移8位


• 或者他们


• 移位可变位数


• 创建2个8位数字，代表16位数字的前半部分和后半部分。

• shift one variable 8 bits left
• OR them
• shift a variable number of bits
• create 2 8 bit numbers representing the first and second half of the 16 bit number.

javascript（根据 https ：//developer.mozilla.org/en/JavaScript/Reference/Operators/Bitwise_Operators ）在转移时使用大端表示...

javascript (according to https://developer.mozilla.org/en/JavaScript/Reference/Operators/Bitwise_Operators) uses big endian representation when shifting...

endianness是一个对我来说有点陌生（我只有90％确定我概述的步骤是我想要的。）所以交换有点令人眼花缭乱。请帮忙！我真的只需要知道如何以有效的方式交换订单。 （我只能想到在toString（）返回值上使用for循环）

endianness is a bit foreign to me (I am only 90 percent sure that my outlined steps are what i want.) so swapping is a bit dizzying. please help! I only really need to know how to swap the order in an efficient manner. (I can only think of using a for loop on a toString() return value)

推荐答案

function swap16(val) {
    return ((val & 0xFF) << 8)
           | ((val >> 8) & 0xFF);
}

说明：

1. 假设 val ，例如， 0xAABB 。


2. 掩码 val 通过 & ing ， 0xFF ：结果是 0xBB 。


3. 转移结果是左边8位：结果是 0xBB00 。


4. Shift val 右边8位：结果是 0xAA （LSB已经掉落了右侧）。


5. 通过 & ing 0xFF ：结果是 0xAA 。


6. 合并结果来自步骤3和步骤5的 | ing 他们在一起：
   
   0xBB00 | 0xAA 是 0xBBAA 。

1. Let's say that val is, for example, 0xAABB.
2. Mask val to get the LSB by &ing with 0xFF: result is 0xBB.
3. Shift that result 8 bits to the left: result is 0xBB00.
4. Shift val 8 bits to the right: result is 0xAA (the LSB has "dropped off" the right-hand side).
5. Mask that result to get the LSB by &ing with 0xFF: result is 0xAA.
6. Combine the results from steps 3 and step 5 by |ing them together:
   0xBB00 | 0xAA is 0xBBAA.

---

---

function swap32(val) {
    return ((val & 0xFF) << 24)
           | ((val & 0xFF00) << 8)
           | ((val >> 8) & 0xFF00)
           | ((val >> 24) & 0xFF);
}

说明：

1. 假设 val ，例如， 0xAABBCCDD 。


2. 掩码 val 通过 & ing ， 0xFF ：结果是 0xDD 。


3. 转移结果是左边24位：结果是 0xDD000000 。


4. 面具 val 通过 & ing <获取第二个字节/ a>与 0xFF00 ：结果是 0xCC00 。


5. Shift ，向左生成8位：结果为 0xCC0000 。


6. Shift val 右边8位：结果是 0xAABBCC （LSB已经掉落右侧）。


7. 通过 & ing ， 0xFF00 ：结果是 0xBB00 。


8. Shift val 右边24位：结果是 0xAA （除了MSB之外的所有内容）已经放弃右侧了。）


9. 通过 & ing ， 0xFF ：结果是 0xAA 。


10. 通过 | ing 他们在一起：
    
    0xDD000000 | 0xCC0000 | 0xBB00 | 0xAA 是 0xDDCCBBAA 。

1. Let's say that val is, for example, 0xAABBCCDD.
2. Mask val to get the LSB by &ing with 0xFF: result is 0xDD.
3. Shift that result 24 bits to the left: result is 0xDD000000.
4. Mask val to get the second byte by &ing with 0xFF00: result is 0xCC00.
5. Shift that result 8 bits to the left: result is 0xCC0000.
6. Shift val 8 bits to the right: result is 0xAABBCC (the LSB has "dropped off" the right-hand side).
7. Mask that result to get the second byte by &ing with 0xFF00: result is 0xBB00.
8. Shift val 24 bits to the right: result is 0xAA (everything except the MSB has "dropped off" the right-hand side).
9. Mask that result to get the LSB by &ing with 0xFF: result is 0xAA.
10. Combine the results from steps 3, 5, 7 and 9 by |ing them together:
    0xDD000000 | 0xCC0000 | 0xBB00 | 0xAA is 0xDDCCBBAA.

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