JavaScript Bitwise Masking

问题

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

The following 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));

Returns:

ef
be00
ad0000
-22000000
ef
beef
adbeef
-21524111

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

ef
be00
ad0000
de000000
ef
beef
adbeef
deadbeef

What's going on with this?

Thank you in advance for your help.

---

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

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

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. 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')));

Which yields the expected output:

ef
be00
ad0000
de000000
ef
beef
adbeef
deadbeef

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

回答1:

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.

One possible fix:

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

回答2:

Because bitwise operations (like &) are done on signed 32-bit integers in javascript. 0xFFFFFFFF, where all bits are set, is actually -1, and the 0xde000000 you are expecting is actually -570425344.

来源：https://stackoverflow.com/questions/28519708/javascript-bitwise-masking