base64のEncodeでのbit演算について
ここのjavasciptのbase64 encodeの実装(下記のコード)でbit演算をしていますが、
var n = (s.charCodeAt(c) << 16) + (s.charCodeAt(c+1) << 8) + s.charCodeAt(c+2);
(1)
n = [(n >>> 18) & 63, (n >>> 12) & 63, (n >>> 6) & 63, n & 63];
(2)
なぜ、(1)と(2)で、base64charsのindexに変換できるのでしょうか?
(1)と(2)にどのような意味があるのでしょうか?
function base64_encode (s)
{
// the result/encoded string, the padding string, and the pad count
var r = "";
var p = "";
var c = s.length % 3;
// add a right zero pad to make this string a multiple of 3 characters
if (c > 0) {
for (; c < 3; c++) {
p += '=';
s += "\0";
}
}
// increment over the length of the string, three characters at a time
for (c = 0; c < s.length; c += 3) {
// we add newlines after every 76 output characters, according to the MIME specs
if (c > 0 && (c / 3 * 4) % 76 == 0) {
r += "\r\n";
}
// these three 8-bit (ASCII) characters become one 24-bit number
var n = (s.charCodeAt(c) << 16) + (s.charCodeAt(c+1) << 8) + s.charCodeAt(c+2); //(1)
// this 24-bit number gets separated into four 6-bit numbers
n = [(n >>> 18) & 63, (n >>> 12) & 63, (n >>> 6) & 63, n & 63]; // (2)
// those four 6-bit numbers are used as indices into the base64 character list
r += base64chars[n[0]] + base64chars[n[1]] + base64chars[n[2]] + base64chars[n[3]];
}
// add the actual padding string, after removing the zero pad
return r.substring(0, r.length - p.length) + p;
}