我实际上能够使用链接的iOS代码来创建以下功能.我没有意识到包含例如单个表情符号的字符串将具有2的长度.因此,您可以检查字符是否实际上是代理.

我不完全确定如何从iOS代码中处理if(substring.length> 1),但我认为Character.isHighSurrogate(myChar)在该实例中执行相同的工作.

private boolean containsIllegalCharacters(String displayName)

{

final int nameLength = displayName.length();

for (int i = 0; i < nameLength; i++)

{

final char hs = displayName.charAt(i);

if (0xd800 <= hs && hs <= 0xdbff)

{

final char ls = displayName.charAt(i + 1);

final int uc = ((hs - 0xd800) * 0x400) + (ls - 0xdc00) + 0x10000;

if (0x1d000 <= uc && uc <= 0x1f77f)

{

return true;

}

}

else if (Character.isHighSurrogate(hs))

{

final char ls = displayName.charAt(i + 1);

if (ls == 0x20e3)

{

return true;

}

}

else

{

// non surrogate

if (0x2100 <= hs && hs <= 0x27ff)

{

return true;

}

else if (0x2B05 <= hs && hs <= 0x2b07)

{

return true;

}

else if (0x2934 <= hs && hs <= 0x2935)

{

return true;

}

else if (0x3297 <= hs && hs <= 0x3299)

{

return true;

}

else if (hs == 0xa9 || hs == 0xae || hs == 0x303d || hs == 0x3030 || hs == 0x2b55 || hs == 0x2b1c || hs == 0x2b1b || hs == 0x2b50)

{

return true;

}

}

}

return false;

}