4 ms·
No, we mean the power operator, 2 to the power of 16 would be '0' in a short int, since the value overflows... But since philosophically a signed int has '1's
by NikkiA 7y ago
No, we mean the power operator, 2 to the power of 16 would be '0' in a short int, since the value overflows...
But since philosophically a signed int has '1's in every bit from the MSB to infinity towards the left, it should thus be -0 not 0.
- leereeves 7y agoIn two's complement (the usual binary representation of signed ints), -0 is the same as 0.