This seems too trivial I guess. But A*B is the one that could overflow. You could do the following, without losing precision A*B – C*D = A(D+E) – (A+F)D = AD + AE – AD – DF = AE – DF ^smaller quantities E & F E = B – D (hence, far smaller than … Read more
Python works: >>> sum(x for x in xrange(1000000000 + 1)) 500000000500000000 Or: >>> sum(xrange(1000000000+1)) 500000000500000000 Python’s int auto promotes to a Python long which supports arbitrary precision. It will produce the correct answer on 32 or 64 bit platforms. This can be seen by raising 2 to a power far greater than the bit width … Read more
-2147483648 is not a “number”. C++ language does not support negative literal values. -2147483648 is actually an expression: a positive literal value 2147483648 with unary – operator in front of it. Value 2147483648 is apparently too large for the positive side of int range on your platform. If type long int had greater range on … Read more
The historical reason is that most C implementations (compilers) just used whatever overflow behaviour was easiest to implement with the integer representation it used. C implementations usually used the same representation used by the CPU – so the overflow behavior followed from the integer representation used by the CPU. In practice, it is only the … Read more
If it overflows, it goes back to the minimum value and continues from there. If it underflows, it goes back to the maximum value and continues from there. You can check that beforehand as follows: public static boolean willAdditionOverflow(int left, int right) { if (right < 0 && right != Integer.MIN_VALUE) { return willSubtractionOverflow(left, -right); … Read more
I see you’re using unsigned integers. By definition, in C (I don’t know about C++), unsigned arithmetic does not overflow … so, at least for C, your point is moot 🙂 With signed integers, once there has been overflow, undefined behaviour (UB) has occurred and your program can do anything (for example: render tests inconclusive). … Read more