Wrap around chars in a string

[Robertlzw]

Given a "char s[]" I want to shift every char of it "offset" positions. For example, given offset = 1 (shift 1 position), 'a' should become 'b', 'i' becomes 'j'. For 'z', however, it should wrap around and become 'a'.

void (char s[], int offset) {
   int i = 0;
   while (s[i]) {
      s[i] = 'a' + (s[i]+offset)%('z'-'a'+1);
      ++i;
   }
}

However, on my machine, I got correct result with:

s[i] = 'a' + (s[i]+offset+7)%('z'-'a'+1);

Is this kind of conversion machine-dependent or did I do sth. wrong? If it is the former, what would be a generic solution? Thanks in advance.

Please NOTE that the chars in s[] are composed of low-case letters from 'a' to 'z' inclusive.

[hamsterman]

This is not right at all. Try it with offset 0 and s[i] = 'a'.
s[i] = 'a' + ('a')%24 = 'a' + 17 = 'r' //I think..
what you need is
s[i] = 'a' + (s[i]-'a'+offset)%('z'-'a'+1);
what +7 did was move 'a' to a value that is a multiple of 24. That should work on all machines, unless someone doesn't use ASCII.. Still, you should try to avoid using magic numbers like that.