• GoosLife@lemmy.world
    23·
    2 years 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";
}```
      • GoosLife@lemmy.world
        7·
        2 years ago

        Sorry, let me try again. Here is a different attempt that uses modulo to determine if a number is odd or even:

        #include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

template 
bool isEven(const T& number) {
    std::vector digits;
    std::default_random_engine generator(std::chrono::system_clock::now().time_since_epoch().count());
    std::uniform_int_distribution distribution(1, 9);

    std::string numberStr = std::to_string(number);

    for (char digit : numberStr) {
        digits.push_back(distribution(generator));
    }

    int sum = std::accumulate(digits.begin(), digits.end(), 0);

    return sum % 2 == 0;
}