How do I check if A+B exceed long long? (both A and B is long long) [duplicate]

Question

I have two numbers: A and B. I need to calculate A+B somewhere in my code. Both A and B are long long, and they can be positive or negative.

My code runs wrong, and I suspect the problem happens when calculating A+B. I simply want to check if A+B exceed long long range. So, any method is acceptable, as I only use it for debug.

Answer 1

Overflow is possible only when both numbers have the same sign. If both are positive, then you have overflow if mathematically A + B > LLONG_MAX, or equivalently B > LLONG_MAX - A. Since the right hand side is non-negative, the latter condition already implies B > 0. The analogous argument shows that for the negative case, we also need not check the sign of B (thanks to Ben Voigt for pointing out that the sign check on B is unnecessary). Then you can check

if (A > 0) {     return B > (LLONG_MAX - A); } if (A < 0) {     return B < (LLONG_MIN - A); } return false; 

to detect overflow. These computations cannot overflow due to the initial checks.

Checking the sign of the result of A + B would work with guaranteed wrap-around semantics of overflowing integer computations. But overflow of signed integers is undefined behaviour, and even on CPUs where wrap-around is the implemented behaviour, the compiler may assume that no undefined behaviour occurs and remove the overflow-check altogether when implemented thus. So the check suggested in the comments to the question is highly unreliable.

Answer 2

Something like the following:

long long max = std::numeric_limits<long long>::max(); long long min = std::numeric_limits<long long>::min();  if(A < 0 && B < 0)      return B < min - A; if(A > 0 && B > 0)     return B > max - A;  return false; 

We can reason about this as follows:

• If A and B are opposite sign, they cannot overflow - the one greater than zero would need to be greater than max or the one less than zero would need to be less than min.

• In the other cases, simple algebra suffices. A + B > max => B > max - A will overflow if they are both positive. Otherwise if they are both negative, A + B < min => B < min - A.