JavaScript的按位屏蔽 [英] JavaScript Bitwise Masking

问题描述

此问题是类似此另一个问题;不过，我想明白为什么这是工作，因为它是。

This question is similar to this other question; however, I'd like to understand why this is working as it is.

以下code：

console.log((parseInt('0xdeadbeef', 16) & parseInt('0x000000ff', 16)).toString(16));
console.log((parseInt('0xdeadbeef', 16) & parseInt('0x0000ff00', 16)).toString(16));
console.log((parseInt('0xdeadbeef', 16) & parseInt('0x00ff0000', 16)).toString(16));
console.log((parseInt('0xdeadbeef', 16) & parseInt('0xff000000', 16)).toString(16));
console.log((parseInt('0xdeadbeef', 16) & parseInt('0x000000ff', 16)).toString(16));
console.log((parseInt('0xdeadbeef', 16) & parseInt('0x0000ffff', 16)).toString(16));
console.log((parseInt('0xdeadbeef', 16) & parseInt('0x00ffffff', 16)).toString(16));
console.log((parseInt('0xdeadbeef', 16) & parseInt('0xffffffff', 16)).toString(16));

返回：

ef
be00
ad0000
-22000000
ef
beef
adbeef
-21524111

当什么，我从.string期待（16）将是：

When what I am expecting from .string(16) would be:

ef
be00
ad0000
de000000
ef
beef
adbeef
deadbeef

这是怎么回事这个？

What's going on with this?

感谢您提前对您有所帮助。

Thank you in advance for your help.

随着谢谢你的反应和评论者下方，同时也为以下来源：

With thank you to the responders and commenters below, and also to the following sources:

• http://speakingjs.com/es5/ch11.html


• JavaScript演员从符号到无符号
空调风格类型

• http://speakingjs.com/es5/ch11.html
• JavaScript C Style Type Cast From Signed To Unsigned

下面是通过提供实用功能的工作原理转换基数为16的32位数字，并从一个符号的32位整数一个解决方案：

Here is a solution that works by providing utility functions to convert a radix-16 32-bit number to and from a signed 32-bit integer:

// Convert 'x' to a signed 32-bit integer treating 'x' as a radix-16 number
// c.f. http://speakingjs.com/es5/ch11.html
function toInt32Radix16(x) {
    return (parseInt(x, 16) | 0);
}

// Convert a signed 32-bit integer 'x' to a radix-16 number
// c.f. http://stackoverflow.com/questions/14890994/javascript-c-style-type-cast-from-signed-to-unsigned
function toRadix16int32(x) {
    return ((x >>> 0).toString(16));
}

console.log(toRadix16int32(toInt32Radix16('0xdeadbeef') & toInt32Radix16('0x000000ff')));
console.log(toRadix16int32(toInt32Radix16('0xdeadbeef') & toInt32Radix16('0x0000ff00')));
console.log(toRadix16int32(toInt32Radix16('0xdeadbeef') & toInt32Radix16('0x00ff0000')));
console.log(toRadix16int32(toInt32Radix16('0xdeadbeef') & toInt32Radix16('0xff000000')));
console.log(toRadix16int32(toInt32Radix16('0xdeadbeef') & toInt32Radix16('0x000000ff')));
console.log(toRadix16int32(toInt32Radix16('0xdeadbeef') & toInt32Radix16('0x0000ffff')));
console.log(toRadix16int32(toInt32Radix16('0xdeadbeef') & toInt32Radix16('0x00ffffff')));
console.log(toRadix16int32(toInt32Radix16('0xdeadbeef') & toInt32Radix16('0xffffffff')));

这产生预期的输出结果：

Which yields the expected output:

ef
be00
ad0000
de000000
ef
beef
adbeef
deadbeef

，以及对我的关于JavaScript的整数行为的一部分了一些很好的学习。

As well as some good learning on my part about JavaScript integer behavior.

推荐答案

在JavaScript中，所有的位操作（和＆安培; 其中）的回报签署的32位整数，结果，范围-2 ³¹ 通过2 ³¹ -1，包容性。这就是为什么你有额外的位（ 0xde000000 大于 0x7ffffff ）再$ P $现在psenting的标志，意义你得到一个负值来代替。

In JavaScript, all bitwise operations (and & among them) return signed 32-bit integer as a result, in the range −2³¹ through 2³¹−1, inclusive. That's why you have that extra bit (0xde000000 is greater than 0x7ffffff) representing a sign now, meaning that you get a negative value instead.

一个可能的修正：

var r = 0xdeadbeef & 0xff000000;
if (r < 0) {
  r += (1 << 30) * 4;
}
console.log( r.toString(16) ); // 'de000000'

这篇关于JavaScript的按位屏蔽的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！