Question

我有一个字符串，它包含UTF-32（但可能更高的16位将始终为0）代码点。每个标记是长字符串中每个字符的代码点的4个字节中的1个。请注意，在变成字符串之前，字节会被解释为signed ints，我无法控制它。

    // Provided: 
    intEncodedBytesString= "0,0,0,-31,0,0,0,-15,0,0,0,-31"; //3 chars: áñá

    // Wanted
    actualCodePoints = [225,241,225];

我需要将intEncodedBytesString转换为actualCodePoints数组。到目前为止，我想出了这个：

var intEncodedBytesStringArray = intEncodedBytesString.toString().split(',');
var i, str = '';
var charAmount = intEncodedBytesStringArray.length / 4;

for (i = 0; i < charAmount; i++) {
  var codePoint = 0;

  for (var j = 0; j < 4; j++) {
    var num = parseInt(intEncodedBytesStringArray[i * 4 + j], 10);
    if (num != 0) {
      if (num < 0) {
        num = (1 << (8 * (4 - j))) + num;
      }

      codePoint += (num << (8 * (3 - j)));
    }
  }

  str += String.fromCodePoint(codePoint);
}

有更好，更简单和/或更有效的方法吗？

我已经看过几十个答案和代码snipets来处理类似的事情，但没有解决我的输入字节在一串签名的int中的问题：S

编辑：此代码不能使用最高代码点，因为1＆lt;＆lt; 32是1而不是2 ^ 32。

Answer 1

因为这是一个很好的简单UTF-32，是的，有一种更简单的方法：只需要工作在四字节块中。此外，处理可能的否定性的简单方法是(value + 256) % 256。

所以：

var intEncodedBytesString = "0,0,0,-31,0,0,0,-15,0,0,0,-31"; //3 char
var actualCodePoints = [];
var bytes = intEncodedBytesString.split(",").map(Number);
for (var i = 0; i < bytes.length; i += 4) {
  actualCodePoints.push(
       (((bytes[i]     + 256) % 256) << 24) +
       (((bytes[i + 1] + 256) % 256) << 16) +
       (((bytes[i + 2] + 256) % 256) << 8) +
       (bytes[i + 3]   + 256) % 256
  );
}

评论中详细解释的示例：

// Starting point
var intEncodedBytesString = "0,0,0,-31,0,0,0,-15,0,0,0,-31"; //3 char
// Target array
var actualCodePoints = [];
// Get the bytes as numbers by splitting on comman running the array
// through Number to convert to number.
var bytes = intEncodedBytesString.split(",").map(Number);

// Loop through the bytes building code points
var i, cp;
for (i = 0; i < bytes.length; i += 4) {
  // (x + 256) % 256 will handle turning (for instance) -31 into 224
  // We shift the value for the first byte left 24 bits, the next byte 16 bits,
  // the next 8 bits, and don't shift the last one at all. Adding them all
  // together gives us the code point, which we push into the array.
  cp = (((bytes[i]     + 256) % 256) << 24) +
       (((bytes[i + 1] + 256) % 256) << 16) +
       (((bytes[i + 2] + 256) % 256) << 8) +
       (bytes[i + 3]   + 256) % 256;
  actualCodePoints.push(cp);
}

// Show the result
console.log(actualCodePoints);

// If the JavaScript engine supports it, show the string
if (String.fromCodePoint) { // ES2015+
  var str = String.fromCodePoint.apply(String, actualCodePoints);
  // The above could be
  // `let str = String.fromCodePoint(...actualCodePoints);`
  // on an ES2015+ engine
  console.log(str);
} else {
  console.log("(Your browser doesn't support String.fromCodePoint)");
}

JavaScript：如何将多字节字符串数组转换为32位int数组？

1 个答案: