Comment on I wish
GoosLife@lemmy.world 1 year ago
There is absolutely no need to add a check for each individual number, just do this:
`#include #include
int main() { int number = 0; int numberToAdd = 1; int modifier = 1;
std::cout << "Is your number [p]ositive or [n]egative? (Default: positive)\n"; if (std::cin.get() == 'n') { modifier *= -1; } std::cin.ignore(std::numeric_limits::max(), '\n'); // Clear the input buffer bool isEven = true; bool running = true; while (running) { std::cout << number << " is " << (isEven ? "even" : "odd") << ".\n"; std::cout << "Continue? [y/n] (Default: yes)\n"; if (std::cin.peek() == 'n') { running = false; } number += numberToAdd * modifier; isEven = !isEven; std::cin.ignore(std::numeric_limits::max(), '\n'); } std::cout << "Your number, " << number << " was " << (isEven ? "even" : "odd") << ".\n";
}`
trashgirlfriend@lemmy.world 1 year ago
I hate this
GoosLife@lemmy.world 1 year ago
Sorry, let me try again. Here is a different attempt that uses modulo to determine if a number is odd or even:
nogrub@lemmy.world 1 year ago
i think you forgot an include there also a goto statement would work better