ЗДЕСЬ Полилиния Кодировка: JavaScript -> Swift
Я пытаюсь реализовать алгоритм кодирования полилинии Javascript ЗДЕСЬ (см. Ниже) в Swift. Я искал в Интернете и не нашел Swift-версию этого алгоритма.
function hereEncodeFloat(value) {
var ENCODING_CHARS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-_';
var result = [];
// convert to fixed point
var fixedPoint = Math.round(value * 100000);
// make room on the lowest bit
fixedPoint = fixedPoint << 1;
// flip bits of negative numbers and ensure that the last bit is set
// (should actually always be the case, but for readability it is ok to do it explicitly)
if (fixedPoint > 0) {
fixedPoint = ~(fixedPoint) | 0x01
}
// var-length encode the number in chunks of 5 bits starting with the least significant
// to the most significant
while (fixedPoint > 0x1F) {
result.push(ENCODING_CHARS[(fixedPoint & 0x1F) | 0x20]);
fixedPoint >>= 5;
}
result.push(ENCODING_CHARS[fixedPoint]);
return result.join('');
}
Есть кто-то, кто может помочь преобразовать это в Swift?
Подробности алгоритма можно найти здесь:
https://developer.here.com/documentation/places/topics/location-contexts.html
Заранее спасибо за помощь,
Джейсон
1 ответ
Решение
Я понял:
func hereEncodeNumber(_ value: Double) -> [Character] {
let ENCODING_CHARS : [Character] = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z","a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","0","1","2","3","4","5","6","7","8","9","-","_"]
var result : [Character] = []
// Convert value to fixed point
let fixedPoint = (value * 100000).rounded(.toNearestOrAwayFromZero)
// Convert fixed point to binary
var binaryNum = Int32(exactly: fixedPoint)!
// Make room on lowest bit
binaryNum = binaryNum << 1
// Flip bits of negative numbers and ensure that last bit is set
// (should actually always be case, but for readability it is ok to do it explicitly)
if binaryNum < 0 {
binaryNum = ~(binaryNum) | 0x01
}
// Var-length encode number in chunks of 5 bits starting with least significant
// to most significant
while binaryNum > 0x1F {
result.append(ENCODING_CHARS[Int((binaryNum & 0x1F) | 0x20)])
binaryNum >>= 5
}
result.append(ENCODING_CHARS[Int(binaryNum)])
return result
}