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    <p>Javascript represents all numbers as 64-bit <a href="http://en.wikipedia.org/wiki/IEEE_754-1985#Double-precision_64_bit" rel="noreferrer">double precision IEEE 754 floating point numbers</a> (see the <a href="http://www.ecma-international.org/publications/files/ECMA-ST/ECMA-262.pdf" rel="noreferrer">ECMAscript spec</a>, 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 Javascript -- the native number data type clearly can't precisely represent a 64-bit int.</p> <p>The following illustrates this. Although javascript <em>appears</em> 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:</p> <pre><code>&lt;html&gt; &lt;head&gt; &lt;script language="javascript"&gt; 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; } &lt;/script&gt; &lt;/head&gt; &lt;body onload="showPrecisionLimits()"&gt; &lt;p&gt;(2^50+1) - (2^50) = &lt;span id="r50"&gt;&lt;/span&gt;&lt;/p&gt; &lt;p&gt;(2^51+1) - (2^51) = &lt;span id="r51"&gt;&lt;/span&gt;&lt;/p&gt; &lt;p&gt;(2^52+1) - (2^52) = &lt;span id="r52"&gt;&lt;/span&gt;&lt;/p&gt; &lt;p&gt;(2^53+1) - (2^53) = &lt;span id="r53"&gt;&lt;/span&gt;&lt;/p&gt; &lt;p&gt;(2^54+1) - (2^54) = &lt;span id="r54"&gt;&lt;/span&gt;&lt;/p&gt; &lt;/body&gt; &lt;/html&gt; </code></pre> <p>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.</p> <pre><code>(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 </code></pre> <hr> <p>So what can you do?</p> <p>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'.</p> <p>For example:</p> <pre><code>var a = [ 0x0000ffff, 0xffff0000 ]; var b = [ 0x00ffff00, 0x00ffff00 ]; var c = [ a[0] &amp; b[0], a[1] &amp; b[1] ]; document.body.innerHTML = c[0].toString(16) + ":" + c[1].toString(16); </code></pre> <p>gets you:</p> <pre><code>ff00:ff0000 </code></pre>
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    1. PObitwise AND in Javascript with a 64 bit integer
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    1. USMichael Gaskill
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    1. PObitwise AND in Javascript with a 64 bit integer
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    1. COThanks. In this case I am actually reading in a binary string containing a 64-bit value. So I could somehow turn that into two 32 bit numbers and use my own internal representation to manage that data.
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      1. USToby Hede
    2. COHey Toby; what do you mean by binary string? If this is a sequence of characters, each of which is the character equivalent of an 8-bit byte, you could do: var a = [ s.charCodeAt(0) + (s.charCodeAt(1) << 8) + (s.charCodeAt(2) << 16) + (s.charCodeAt(3) << 24), s.charCodeAt(4) + (s.charCodeAt(5) << 8) + (s.charCodeAt(6) << 16) + (s.charCodeAt(7) << 24) ]; Just keep an eye on the endian-ness of things.
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    3. CO@Orent Trutner: be careful here: with Unicode, a char code could exceed 255. I think that your code fails as soon as one of the bytes has the high bit set.
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