1. i = (unsigned short)65535 + (unsigned short)3; 2. i = (unsigned short)1 - (unsigned short)3; 3. li = (unsigned long)0xFFFFFFFF + (unsigned long)3; 4. li = (unsigned long)1 - (unsigned long)3; TIA Andy Nov 14 '05, 12:33 AM Re: unsigned short addition/subtraction ove...