字符编码独立字符交换

Question

I like to use this piece of code when I want to reverse a string. _{[When I am not using std::string or other inbuilt functions in C ]} . As a beginner when I initially thought of this I had ASCII table in mind. I think this can work well with Unicode too. I assumed since the difference in values (ASCII etc) is fixed, so it works.

Are there any character encodings in which this code may not work?

char a[11],t;
int len,i;
strcpy(a,"Particl");    
printf("%s\n",a);
len = strlen(a);
for(i=0;i<(len/2);i++)
{
    a[i] += a[len-1-i];
    a[len-1-i] = a[i] - a[len-1-i];
    a[i] -= a[len-1-i];
}
printf("%s\n",a);

_Update:

_{This link is informative in association with this question.}

Answer 1

This will not work with any encoding in which some (not necessarily all) codepoints require more than one char unit to represent, because you are reversing byte-by-byte instead of codepoint-by-codepoint. For the usual 8-bit char this includes all encodings that can represent all of Unicode.

For example: in UTF-16BE, the string "hello" maps to the byte sequence 00 68 00 65 00 6c 00 6c 00 6f . Your algorithm applied to this byte sequence will produce the sequence 6f 00 6c 00 6c 00 65 00 68 00 , which is the UTF-16BE encoding of the string "漀氀氀攀栀".

It gets worse -- doing a codepoint-by-codepoint reversal of a Unicode string still won't produce the correct results in all cases, because Unicode has many codepoints that act on their surroundings rather than standing alone as characters. As a trivial example, codepoint-reversing the string "Spın̈al Tap", which contains U+0308 COMBINING DIAERESIS, will produce "paT länıpS" -- see how the diaeresis has migrated from the N to the A? The consequences of codepoint-by-codepoint reversal on a string containing bidirectional overrides or conjoining jamo would be even more dire.