Subversion Repositories Applications.papyrus

if(!dojo._hasResource["dojox.encoding.easy64"]){ //_hasResource checks added by build. Do not use _hasResource directly in your code.
dojo._hasResource["dojox.encoding.easy64"] = true;
dojo.provide("dojox.encoding.easy64");

(function(){
        var c = function(input, length, result){
                for(var i = 0; i < length; i += 3){
                        result.push(
                                String.fromCharCode((input[i] >>> 2) + 33),
                                String.fromCharCode(((input[i] & 3) << 4) + (input[i + 1] >>> 4) + 33),
                                String.fromCharCode(((input[i + 1] & 15) << 2) + (input[i + 2] >>> 6) + 33),
                                String.fromCharCode((input[i + 2] & 63) + 33)
                        );
                }
        };
        
        dojox.encoding.easy64.encode = function(input){
                // summary: encodes input data in easy64 string
                // input: Array: an array of numbers (0-255) to encode
                var result = [], reminder = input.length % 3, length = input.length - reminder;
                c(input, length, result);
                if(reminder){
                        var t = input.slice(length);
                        while(t.length < 3){ t.push(0); }
                        c(t, 3, result);
                        for(var i = 3; i > reminder; result.pop(), --i);
                }
                return result.join(""); // String
        };

        dojox.encoding.easy64.decode = function(input){
                // summary: decodes the input string back to array of numbers
                // input: String: the input string to decode
                var n = input.length, r = [], b = [0, 0, 0, 0], i, j, d;
                for(i = 0; i < n; i += 4){
                        for(j = 0; j < 4; ++j){ b[j] = input.charCodeAt(i + j) - 33; }
                        d = n - i;
                        for(j = d; j < 4; b[++j] = 0);
                        r.push(
                                (b[0] << 2) + (b[1] >>> 4),
                                ((b[1] & 15) << 4) + (b[2] >>> 2),
                                ((b[2] & 3) << 6) + b[3]
                        );
                        for(j = d; j < 4; ++j, r.pop());
                }
                return r;
        };
})();

}