按位与在Javascript中有64位整数 [英] bitwise AND in Javascript with a 64 bit integer
问题描述
我要寻找的进行按位和64位整数的JavaScript的方式。
JavaScript的会投它的所有double值到符号的32位整数做位操作（<一href="https://developer.mozilla.org/En/Core_JavaScript_1.5_Reference/Operators/Bitwise_Operators">details这里）。
Java脚本重新presents所有号码为64位的双击precision IEEE 754浮点数（见的的ECMAScript规范，8.5节。）所有的正整数，达到2 ^ 53可连接codeD precisely。较大的整数让他们至少显著位裁剪。显然本机号码数据类型不能precisely重新present 64位int
下面说明了这一点。虽然JavaScript的出现的，以便能够解析进制数再presenting 64位数字，底层数字再presentation未持有64位。尝试在浏览器中执行以下操作：
＆LT; HTML＆GT;
＆LT; HEAD＆GT;
＆LT; SCRIPT LANGUAGE =JavaScript的＆GT;
功能显示precisionLimits（）{
的document.getElementById（R50）的innerHTML = 0x0004000000000001 - 0x0004000000000000。
的document.getElementById（R51）的innerHTML = 0x0008000000000001 - 0x0008000000000000。
的document.getElementById（R52）的innerHTML = 0x0010000000000001 - 0x0010000000000000。
的document.getElementById（R53）的innerHTML = 0x0020000000000001 - 0x0020000000000000。
的document.getElementById（R54）的innerHTML = 0x0040000000000001 - 0x0040000000000000。
}
＆LT; / SCRIPT＆GT;
＆LT; /头＆GT;
＆LT;身体的onload =显示precisionLimits（）＆GT;
＆其中p为H.;（2 ^ 50 + 1） - （2 ^ 50）=＆其中;跨度的id =R50＆GT;＆所述; /跨度＆GT;＆所述; / P＆GT;
＆其中p为H.;（2 ^ 51 + 1） - （2 ^ 51）=＆其中;跨度的id =R51＆GT;＆所述; /跨度＆GT;＆所述; / P＆GT;
＆其中p为H.;（2 ^ 52 + 1） - （2 ^ 52）=＆其中;跨度的id =R52＆GT;＆所述; /跨度＆GT;＆所述; / P＆GT;
＆其中p为H.;（2 ^ 53 + 1） - （2 ^ 53）=＆其中;跨度的id =R53＆GT;＆所述; /跨度＆GT;＆所述; / P＆GT;
＆其中p为H.;（2 ^ 54 + 1） - （2 ^ 54）=＆其中;跨度的id =R54＆GT;＆所述; /跨度＆GT;＆所述; / P＆GT;
＆LT; /身体GT;
＆LT; / HTML＆GT;
在Firefox，Chrome和IE浏览器，我发现了以下内容。如果数字被储存在其全部64位的辉煌，其结果应该是1对所有substractions。相反，你可以看到2 ^ 53 + 1和2 ^ 53之间的差异将会丢失。
（2 ^ 50 + 1） - （2 ^ 50）= 1
（2 ^ 51 + 1） - （2 ^ 51）= 1
（2 ^ 52 + 1） - （2 ^ 52）= 1
（2 ^ 53 + 1） - （2 ^ 53）= 0
（2 ^ 54 + 1） - （2 ^ 54）= 0
所以，你该怎么办？
如果您选择重新present 64位整数作为两个32位的数字，然后应用按位与很简单，只要将2位与的，低，高的32位字。
例如：
VAR一个= [0x0000ffff，为0xffff0000]。
变种B = [0x00ffff00，0x00ffff00]。
变种C =〔一个[0]＆安培; B [0]，A [1]＆安培; B〔1]];
document.body.innerHTML = C [0]的ToString（16）+：+ C [1]的ToString（16）;
让你：
FF00：FF0000
I am looking for a way of performing a bitwise AND on a 64 bit integer in JavaScript.
JavaScript will cast all of its double values into signed 32-bit integers to do the bitwise operations (details here).
Javascript represents all numbers as 64-bit double precision IEEE 754 floating point numbers (see the ecmascript spec, section 8.5.) All positive integers up to 2^53 can be encoded precisely. Larger integers get their least significant bits clipped. This leaves the question of how can you even represent a 64-bit integer in javascipt -- the native number data type clearly can't precisely represent a 64-bit int.
The following illustrates this. Although javascript appears to be able to parse hexadecimal numbers representing 64-bit numbers, the underlying numeric representation does not hold 64 bits. Try the following in your browser:
<html>
<head>
<script language="javascript">
function showPrecisionLimits() {
document.getElementById("r50").innerHTML = 0x0004000000000001 - 0x0004000000000000;
document.getElementById("r51").innerHTML = 0x0008000000000001 - 0x0008000000000000;
document.getElementById("r52").innerHTML = 0x0010000000000001 - 0x0010000000000000;
document.getElementById("r53").innerHTML = 0x0020000000000001 - 0x0020000000000000;
document.getElementById("r54").innerHTML = 0x0040000000000001 - 0x0040000000000000;
}
</script>
</head>
<body onload="showPrecisionLimits()">
<p>(2^50+1) - (2^50) = <span id="r50"></span></p>
<p>(2^51+1) - (2^51) = <span id="r51"></span></p>
<p>(2^52+1) - (2^52) = <span id="r52"></span></p>
<p>(2^53+1) - (2^53) = <span id="r53"></span></p>
<p>(2^54+1) - (2^54) = <span id="r54"></span></p>
</body>
</html>
In Firefox, Chrome and IE I'm getting the following. If numbers were stored in their full 64-bit glory, the result should have been 1 for all the substractions. Instead, you can see how the difference between 2^53+1 and 2^53 is lost.
(2^50+1) - (2^50) = 1
(2^51+1) - (2^51) = 1
(2^52+1) - (2^52) = 1
(2^53+1) - (2^53) = 0
(2^54+1) - (2^54) = 0
So what can you do?
If you choose to represent a 64-bit integer as two 32-bit numbers, then applying a bitwise AND is as simple as applying 2 bitwise AND's, to the low and high 32-bit 'words'.
For example:
var a = [ 0x0000ffff, 0xffff0000 ];
var b = [ 0x00ffff00, 0x00ffff00 ];
var c = [ a[0] & b[0], a[1] & b[1] ];
document.body.innerHTML = c[0].toString(16) + ":" + c[1].toString(16);
gets you:
ff00:ff0000
这篇关于按位与在Javascript中有64位整数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！
