Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
630 views
in Technique[技术] by (71.8m points)

sign - Javascript: convert a (hex) signed integer to a javascript value

I have a signed value given as a hex number, by example 0xffeb and want convert it into -21 as a "normal" Javascript integer.

I have written some code so far:

function toBinary(a) { //: String
    var r = '';
    var binCounter = 0;
    while (a > 0) {
        r = a%2 + r;
        a = Math.floor(a/2);
    }
    return r;
}

function twoscompl(a) { //: int
    var l = toBinaryFill(a).length;
    var msb = a >>> (l-1);

    if (msb == 0) {
        return a;
    }

    a = a-1;
    var str = toBinary(a);
    var nstr = '';
    for (var i = 0; i < str.length; i++) {
        nstr += str.charAt(i) == '1' ? '0' : '1';
    }
    return (-1)*parseInt(nstr);
}

The problem is, that my function returns 1 as MSB for both numbers because only at the MSB of the binary representation "string" is looked. And for this case both numbers are 1:

-21 => 0xffeb => 1111 1111 1110 1011
 21 => 0x15   =>              1 0101

Have you any idea to implement this more efficient and nicer?

Greetings, mythbu

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Use parseInt() to convert (which just accepts your hex string):

parseInt(a);

Then use a mask to figure out if the MSB is set:

a & 0x8000

If that returns a nonzero value, you know it is negative.

To wrap it all up:

a = "0xffeb";
a = parseInt(a, 16);
if ((a & 0x8000) > 0) {
   a = a - 0x10000;
}

Note that this only works for 16-bit integers (short in C). If you have a 32-bit integer, you'll need a different mask and subtraction.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...