Example analysis of JavaScript implementation of encode64 encryption algorithm

The example of this article describes the encode64 encryption algorithm implemented by JavaScript. Share with you for your reference. The details are as follows:

This JavaScript code can implement the encode64 encryption algorithm, the speed is quite good.

//encode64 codec 
(function() {
 var codeChar = "PaAwO65goUf7IK2vi9-xq8cFTEXLCDY1Hd3tV0ryzjbpN_BlnSs4mGRkQWMZJeuh";
 window.encode64 = function(str) {
  var s = "";
  var a = strToBytes(str);
  // Gets the byte array of a string ,  The array length is the string length 2 times .
  var res = a.length % 3;
  //3 bytes 1 Group processing ,  Remaining special treatment 
  var i = 2, v;
  for (; i < a.length; i += 3) {
  // every 3 Bytes in 4 Character representation , 
  // The equivalent of 3 A character ( Is, in fact, 6 bytes ) with 8 Character encoding ( For the actual 16 bytes )
  // It looks like the capacity has expanded a lot ,  But with compression enabled ,  These cancel out again 
   v = a[i - 2] + (a[i - 1] << 8) + (a[i] << 16);
   s += codeChar.charAt(v & 0x3f);
   s += codeChar.charAt((v >> 6) & 0x3f);
   s += codeChar.charAt((v >> 12) & 0x3f);
   s += codeChar.charAt((v >> 18));
  }
  if (res == 1) {// The word "savings" 1 When a ,  fill 2 A character , 64*64>256
   v = a[i - 2];
   s += codeChar.charAt(v & 0x3f);
   s += codeChar.charAt((v >> 6) & 0x3f);
  } else if (res == 2) {
  // The word "savings" 2 When a ,  fill 3 bytes , 64*64*64>256*256,  So it works 
   v = a[i - 2] + (a[i - 1] << 8);
   s += codeChar.charAt(v & 0x3f);
   s += codeChar.charAt((v >> 6) & 0x3f);
   s += codeChar.charAt((v >> 12) & 0x3f);
  }
  return s;
 };
 window.decode64 = function(codeStr) {
  var dic = [];
  for (var i = 0; i < codeChar.length; i++) {
   dic[codeChar.charAt(i)] = i;
  }
  var code = [];
  var res = codeStr.length % 4;
  var i = 3, v;
  for (; i < codeStr.length; i += 4) {
   v = dic[codeStr.charAt(i - 3)];
   v += dic[codeStr.charAt(i - 2)] << 6;
   v += dic[codeStr.charAt(i - 1)] << 12;
   v += dic[codeStr.charAt(i)] << 18;
   code.push(v & 0xff, (v >> 8) & 0xff, (v >> 16) & 0xff);
  }
  if (res == 2) {
  // The correct number of bytes must be redundant 2 or 3,  There is no 1 In the case ,  If there is a ,  Give up .
   v = dic[codeStr.charAt(i - 3)];
   v += dic[codeStr.charAt(i - 2)] << 6;
   code.push(v & 0xff);
  } else if (res == 3) {
   v = dic[codeStr.charAt(i - 3)];
   v += dic[codeStr.charAt(i - 2)] << 6;
   v += dic[codeStr.charAt(i - 1)] << 12;
   code.push(v & 0xff, (v >> 8) & 0xff);
  }
  return strFromBytes(code);
 };
})();

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