If you’re interested in a purely theoretical answer, the C++ standard allows undefined behaviour to “time travel”: [intro.execution]/5: A conforming implementation executing a well-formed program shall produce the same observable behavior as one of the possible executions of the corresponding instance of the abstract machine with the same program and the same input. However, if … Read more
This is due to integer overflow. When thisVal is very large and anotherVal is negative then subtracting the latter from the former yields a result that is bigger than thisVal which may overflow to the negative range.
1. Detecting the overflow: x = a * b; if (a != 0 && x / a != b) { // overflow handling } Edit: Fixed division by 0 (thanks Mark!) 2. Computing the carry is quite involved. One approach is to split both operands into half-words, then apply long multiplication to the half-words: uint64_t … Read more
is still overflow of these types an undefined behavior? Yes. Per Paragraph 5/4 of the C++11 Standard (regarding any expression in general): If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined. […] The fact that a … Read more
No, your 2nd code isn’t correct, but you are close: if you set int half = INT_MAX/2; int half1 = half + 1; the result of an addition is INT_MAX. (INT_MAX is always an odd number). So this is valid input. But in your routine you will have INT_MAX – half == half1 and you … Read more
Expanding on user71404’s answer: int f(unsigned x) { if (x <= INT_MAX) return static_cast<int>(x); if (x >= INT_MIN) return static_cast<int>(x – INT_MIN) + INT_MIN; throw x; // Or whatever else you like } If x >= INT_MIN (keep the promotion rules in mind, INT_MIN gets converted to unsigned), then x – INT_MIN <= INT_MAX, so … Read more
If the optimiser does such a reordering it is still bound to the C specification, so such a reordering would become: uint64_t u64_z = (uint64_t)u32_x + (uint64_t)u32_y + u64_a; Rationale: We start with uint64_t u64_z = u32_x + u64_a + u32_y; Addition is performed left-to-right. The integer promotion rules state that in the first addition … Read more
Java 8 has Math.multiplyExact, Math.addExact etc. for ints and long. These throw an unchecked ArithmeticException on overflow.
Here is what the program does at each step: 1 * 10 = 10 10 * 11 = 110 110 * 12 = 1320 1320 * 13 = 17160 17160 * 14 = 240240 240240 * 15 = 3603600 3603600 * 16 = 57657600 57657600 * 17 = 980179200 980179200 * 18 = 463356416 463356416 … Read more
Because 2000 is an int which is usually 32-bit. Just use 2000LL. Using LL suffix instead of ll was suggested by @AdrianMole in, now deleted, comment. Please check his answer. By default, integer literals are of the smallest type that can hold their value but not smaller than int. 2000 can easily be stored in … Read more