C ++语法错误：程序以非零状态退出

Question

我正在使用repl.it编写我的C ++。 到目前为止，我已经了解了条件，循环和函数。 现在，我正在尝试编写一个输入两个整数并找到最小公倍数和最大公分母的程序。 到目前为止，我已经编写了大部分代码，但是存在问题。

“以非零状态退出”

#include <iostream>

using namespace std;

int main() 
{
int number1 = 0;
int number2 = 0;
int calc = 0;
int lcm = 0;
cout << "Give me two integers, and I will calculate the Least Common Multiple and the Greatest Common Divisor." << endl;

while (number1 <= 0) {

    cout << "Enter your first number. Cant be negative" << endl;
    cin >> number1; }

while (number2 <= 0) {
    cout << "Enter your second number. Cant be negative" << endl;
    cin >> number2; }

while(number2 != 0) { ///Greatest Common Divisor
    calc = number1 % number2;
    lcm = (number1*number2) / calc;
    number1 = number2;
    number2 = calc;
}



cout << "Least Common Multiple is " << lcm << endl;

cout << "Greatest Common divisor is " << number1 << endl;
}

所以我不确定这是语法错误还是由于repl.it而引起的，但是我真的很难弄清楚这一点。

谢谢

Answer 1

虽然循环在循环结束时检查number2值，但是当您计算calc时，有时该值为零，然后在下一步中程序被零除异常退出。 您可以通过在计算calc变量后将此行添加到代码中来防止出现此问题：

if (calc == 0 ) break;

另外，您的代码无法正常工作，例如，设置number1 = 30和number2 = 18！

我使用二进制方法计算GCD，然后使用GCD计算LCM。

            #include <iostream>
            #include <math.h>   //  for pow(2,d)
            using namespace std;


            int main() 
            {
                int gcd, lcm, a, b, g, number1 = 0, number2 = 0, d=0;
                cout << "Give me two integers, and I will calculate the Greatest Common Divisor and the Least Common Multiple." << endl;

                while (number1 <= 0) {
                    cout << "Enter your first number. Cant be negative" << endl;
                    cin >> number1; 
                }
                // using binary method to calculating GCD:   https://en.wikipedia.org/wiki/Greatest_common_divisor
                while (number2 <= 0) {
                    cout << "Enter your second number. Cant be negative" << endl;
                    cin >> number2; 
                }    
                a = number1;
                b = number2;
                while (((a%2)==0) && ((b%2)==0)) {
                    a = a/2;
                    b = b/2;
                    d = d+1;
                }
                while (a != b) {
                    if ((a%2) == 0) {
                        a = a/2;
                    } else if ((b%2)==0) {
                        b = b/2;
                    } else if (a>b) {
                        a = (a-b) /2;
                    } else {
                        b = (b-a)/2;
                    }
                }
                g = a;
                cout << "\ng: " << g << "\td: " << d << "\tpower(2,d): " << pow(2,d);
                gcd = g * pow(2,d); // power(2,d) with math.h library
                lcm = (number1*number2)/gcd;   // according to LCM(a,b) = (a*b)/GCD(a,b)
                cout << "\nGreatest Common Divisor is " << gcd << " and Least Common Multiple is " << lcm << endl;
            }

Answer 2

首先，您需要检查哪个数字更大，因为如果您设置number1=18和number2=30那么number1%number2将为18，然后您会看到错误的输出。 此外，Hossein Nazari建议的修复程序可以避免出现异常，但是，如果遵循该修复程序，则GCD将存储在number2 。 例如30和18：

30%18 = 12
calc = 12
if(calc==0) break;
(lcm routine)
number1 = 18
number2 = 12

18%12 = 6
calc=6
if(calc==0) break;
(lcm routine)
number1 = 12
number2 = 6

12%6=0
calc=0
if(calc==0) break;
<done>

我宁愿从您的循环中完全删除lcm计算，在这种情况下，您将按计划将GCD放入number1 ，然后仅通过公式(number1*number2)/GCD计算LCM，即30*18/6 = 90; (number1*number2)/GCD 30*18/6 = 90;

Answer 3

您好，只需更改这两个的数据类型

float calc = 0;
float lcm = 0;

@Cheers和hth，您好，我在这里执行了代码https://www.tutorialspoint.com/compile_cpp11_online.php ，并得到了这个异常-浮点异常，在我将数据类型从int更改为float之后，我得到了这个值输入您的第一个数字。 不能为负
10
输入您的第二个号码。 不能为负
20
最小公倍数为inf
最大公约数是10