Here you can find the source of toAscii85()
String.prototype.toAscii85 = function() { let pad = '\0', addnum = 0, str = this, res = '', sum = 0; while (str.length % 4 !== 0) {/*from w w w .ja v a2 s. c o m*/ str += pad; addnum ++; } for (let i = 0; i < str.length; i += 4) { let e = ''; let temp = str.substr(i, 4); for (let j = 0; j < 4; j++) { let t = temp[j].charCodeAt().toString(2) while (t.length !== 8) t = '0' + t; e += t; } sum = parseInt(e, 2); let encode = []; for (let j = 0; j < 5; j++) encode.push(0); for (let j = 4; j >= 0; j--) { encode[j] = sum % 85; sum = Math.floor(sum / 85); } for (let j = 0; j < 5; j++) res += String.fromCharCode(encode[j] + 33); } res = res.substr(0, res.length - addnum); let newres = '', i = 0; for (i = 0; i < res.length; i += 5) newres += res.substr(i, Math.min(5, res.length - i)) === '!!!!!' ? 'z' : res.substr(i, Math.min(5, res.length - i)); return '<~' + newres + '~>'; };
String.prototype.toAscii85 = function() { var str = this, result = '' while (str.length) { let num = 0, j = 5, i = 4 if (str.substr(0,4) == '\u0000\u0000\u0000\u0000') result += 'z' else { while (i--) { num += (str.charCodeAt(3-i) || 0)*Math.pow(2, 8*i) while (j-- && str.length > 3-j) { result += String.fromCharCode(num/Math.pow(85, j) % 85 + 33) str = str.substr(4) return `<~${result}~>`
String.prototype.toAscii85 = function() { var ret = ""; for(var i=0; i<this.length; i+=4){ var deficit = Math.max(0, 4+i-this.length); var nnn = 0; for(var j=0; j<4; j++){ nnn = nnn*256 + (this.charCodeAt(i+j) || 0); var codes = []; ...
String.prototype.toAscii85 = function() { var str = ''; for(var i = 0, len = this.length; i < len; i+=4) { var bin = ''; var zeroes = 0; for(var j = 0; j < 4; j++) { var char = this[i+j]; bin += to8bit(char ? char.charCodeAt(0) : (zeroes++, 0)); var n = parseInt(bin, 2); for(var j = 4; j >= 0; j--) { var c = Math.floor(n / Math.pow(85, j)); n -= c*Math.pow(85, j); str += String.fromCharCode(c + 33); str = str.slice(0, zeroes > 0 ? -zeroes : undefined); return '<~' + str.replace(/!{5}/g, 'z') + '~>'