Programming Forums - View Single Post

Ooble · Jan 15th, 2005, 9:21 AM

You could always write your own:

int count;
 for (int i = 0; str[i]; i++) {
     if (((str[i] & 0xDF) < 'A') || ((str[i] & 0xDF) > 'Z')) {
         count++;
 }
 count++;

That should do it. I believe it's about as fast as you can get, though there's probably more bit operations you can do. In case you hadn't realised, the str[i] & 0xDF bit converts the character to uppercase - it can be replaced by str[i] - 32 if you like.

EDIT: just realised this was the Java section :o - good luck with the conversion. Unfortunately, I don't know Java, so I can't help with that.

Jan 15th, 2005, 9:21 AM	#4
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	You could always write your own: (Toggle Plain Text) int count; for (int i = 0; str[i]; i++) { if (((str[i] & 0xDF) < 'A') \|\| ((str[i] & 0xDF) > 'Z')) { count++; } count++; int count; for (int i = 0; str[i]; i++) { if (((str[i] & 0xDF) < 'A') \|\| ((str[i] & 0xDF) > 'Z')) { count++; } count++; That should do it. I believe it's about as fast as you can get, though there's probably more bit operations you can do. In case you hadn't realised, the str[i] & 0xDF bit converts the character to uppercase - it can be replaced by str[i] - 32 if you like. EDIT: just realised this was the Java section :o - good luck with the conversion. Unfortunately, I don't know Java, so I can't help with that. __________________ Me :: You :: Them Last edited by Ooble; Jan 15th, 2005 at 9:27 AM.