Question

Using bitwise operators and I suppose addition and subtraction, how can I check if a signed integer is positive (specifically, not negative and not zero)? I’m sure the answer to this is very simple, but it’s just not coming to me.

Answer 1

If you really want an “is strictly positive” predicate for int n without using conditionals (assuming 2’s complement):

• -n will have the sign (top) bit set if n was strictly positive, and clear in all other cases except n == INT_MIN;
• ~n will have the sign bit set if n was strictly positive, or 0, and clear in all other cases including n == INT_MIN;
• …so -n & ~n will have the sign bit set if n was strictly positive, and clear in all other cases.

Apply an unsigned shift to turn this into a 0 / 1 answer:

int strictly_positive = (unsigned)(-n & ~n) >> ((sizeof(int) * CHAR_BIT) - 1);

EDIT: as caf points out in the comments, -n causes an overflow when n == INT_MIN (still assuming 2’s complement). The C standard allows the program to fail in this case (for example, you can enable traps for signed overflow using GCC with the-ftrapv option). Casting n to unsigned fixes the problem (unsigned arithmetic does not cause overflows). So an improvement would be:

unsigned u = (unsigned)n;
int strictly_positive = (-u & ~u) >> ((sizeof(int) * CHAR_BIT) - 1);