Don’t use %
. This is a problem that calls for bitmasks:
bool same_parity = (i & 0x1) == (j & 0x1);
This works regardless of the sign of i
, since the result of that expression will always be 0
or 1
.
Don’t use %
. This is a problem that calls for bitmasks:
bool same_parity = (i & 0x1) == (j & 0x1);
This works regardless of the sign of i
, since the result of that expression will always be 0
or 1
.