JavaScript 按位掩码

这个问题类似于这另一个问题;但是，我想了解为什么会这样。

下面的代码：

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));

Returns:

ef
be00
ad0000
-22000000
ef
beef
adbeef
-21524111

当我对 .string(16) 的期望是：

ef
be00
ad0000
de000000
ef
beef
adbeef
deadbeef

这是怎么回事？

预先感谢您的帮助。

感谢以下回复者和评论者，以及以下来源：

• http://speakingjs.com/es5/ch11.html
• JavaScript C 风格类型从有符号转换为无符号

下面是一个解决方案，它通过提供实用函数来将 radix-16 32 位数字与有符号 32 位整数相互转换：

// 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. https://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')));

这会产生预期的输出：

ef
be00
ad0000
de000000
ef
beef
adbeef
deadbeef

以及我对 JavaScript 整数行为的一些很好的学习。

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.

一种可能的修复：

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