C++中可正确获取UTF-8字符长度的函数分享

2019/7/10 23:15:13

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在C++的char*以及string中,使用的是字节流编码,即sizeof(char) == 1。

也就是说,C++是不区分字符的编码的。

而一个合法UTF8的字符长度可能为1~4位。

现在假设一串输入为UTF8编码,如何能准确的定位到每个UTF8字符的“CharPoint”,而不会错误的分割字符呢?

参考这个页面:http://www.nubaria.com/en/blog/?p=289

可以改造出下面的函数:

const unsigned char kFirstBitMask = 128; // 1000000
const unsigned char kSecondBitMask = 64; // 0100000
const unsigned char kThirdBitMask = 32; // 0010000
const unsigned char kFourthBitMask = 16; // 0001000
const unsigned char kFifthBitMask = 8; // 0000100
 
int utf8_char_len(char firstByte)
{
  std::string::difference_type offset = 1;

  if(firstByte & kFirstBitMask) // This means the first byte has a value greater than 127, and so is beyond the ASCII range.
  {  
    if(firstByte & kThirdBitMask) // This means that the first byte has a value greater than 224, and so it must be at least a three-octet code point.
    {  
      if(firstByte & kFourthBitMask) // This means that the first byte has a value greater than 240, and so it must be a four-octet code point.
        offset = 4;
      else
        offset = 3;
    }  
    else
    {  
      offset = 2;
    }  
  }  
  return offset;
}



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