> IIRC s and ss are pointers to a (the same?) string, right? If they are, > then this difference here will easily fit into 32bit so you could cast > to int explicitly: > > return (int)((s - ss) - 1) > > Or you could change the function definition to return a long instead of > int and return a 64bit value on 64bit systems. Yes, that will do. Thanks! Martin -- Posted via http://www.ruby-forum.com/.