Objective

Implement a function that says whether a + b would leave the range of an int, and says it without ever computing a + b.

Steps

• a and b are the two values whose sum is in question. Return 1 when that sum would not fit in an int, 0 when it would.
• Never evaluate a + b. Signed overflow is undefined behavior, so a check written after the fact, like if (a + b < a), is not slow, it is broken: the compiler is entitled to assume the overflow never happens and to delete the comparison. With optimizations on, that is what it does, and your function starts returning 0 for every input.
• Compare a against the bound instead. If b is positive, the sum runs past the top of the range exactly when a > INT_MAX - b. If b is negative, it runs past the bottom exactly when a < INT_MIN - b. Both subtractions stay inside the range, so neither of them can overflow.
• INT_MAX and INT_MIN come from <limits.h>.
• A b of 0 can never push the sum anywhere: return 0, including for a equal to INT_MIN or INT_MAX.
• Operands of opposite signs never overflow. INT_MAX + INT_MIN is -1, well inside the range, so it returns 0. Only two positives or two negatives can escape.
• The boundary is exact, not approximate. INT_MAX - 1 plus 1 fits and returns 0, while INT_MAX - 1 plus 2 does not and returns 1.
• The tests print your return value followed by a newline, so 1 or 0 on its own line.

Objective

Implement a function that says whether a + b would leave the range of an int, and says it without ever computing a + b.

Steps

• a and b are the two values whose sum is in question. Return 1 when that sum would not fit in an int, 0 when it would.
• Never evaluate a + b. Signed overflow is undefined behavior, so a check written after the fact, like if (a + b < a), is not slow, it is broken: the compiler is entitled to assume the overflow never happens and to delete the comparison. With optimizations on, that is what it does, and your function starts returning 0 for every input.
• Compare a against the bound instead. If b is positive, the sum runs past the top of the range exactly when a > INT_MAX - b. If b is negative, it runs past the bottom exactly when a < INT_MIN - b. Both subtractions stay inside the range, so neither of them can overflow.
• INT_MAX and INT_MIN come from <limits.h>.
• A b of 0 can never push the sum anywhere: return 0, including for a equal to INT_MIN or INT_MAX.
• Operands of opposite signs never overflow. INT_MAX + INT_MIN is -1, well inside the range, so it returns 0. Only two positives or two negatives can escape.
• The boundary is exact, not approximate. INT_MAX - 1 plus 1 fits and returns 0, while INT_MAX - 1 plus 2 does not and returns 1.
• The tests print your return value followed by a newline, so 1 or 0 on its own line.