NSString length with flag emojis

问题

I'm trying to get the length of an NSString that contains a bunch of emojis, including the flag characters. Now I know how to get the length of a string containing emojis:

__block NSInteger length = 0;
[string enumerateSubstringsInRange:range
                           options:NSStringEnumerationByComposedCharacterSequences
                        usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
    length++;
}];

However, "NSStringEnumerationByComposedCharacterSequences" doesn't work in this case. Normal emojis are composed of two characters put together. The flag emojis are composed of two emoji-like characters put together. Thus the length of one flag emoji without any enumeration technique is 4.

How do I get the proper length (number of characters) of an NSString containing flag emojis?

回答1:

I tried the below solution. For normal emoji it returns 2 and for flags it will give 4. Also this will work with normal chars and special chars.

const char *cString = [string UTF8String];

int textLength = (int)strlen(cString);

回答2:

I strongly recommended you read This Article by Ole Begemann, which explains NSString/Unicode/Encoding, alone with talking about some pitfalls about NSString length count, comparison, etc.

According to this article:

You may use:

[text lengthOfBytesUsingEncoding:NSUTF32StringEncoding] / 4

来源：https://stackoverflow.com/questions/32365686/nsstring-length-with-flag-emojis