检查数字

Question

我正在尝试制作一个程序，检查一个数字中的所有数字是否都是奇数，但只有当输入的数字是1位数时才会显示该数字仅由奇数位组成的cout。 这是代码：

int main()
{
    int n, c;
    cin >> n;
    while (n != 0) {
        c = n % 10;
        if (c % 2 == 1) {
            n = n / 10;
            if (n == 0) {
                cout << " Number has only odd digits";
            }
        }
        else
            cout << " Number doesn't have only odd digits";
        return 0;
    }
}

Answer 1

我建议将数字保存为字符串，然后在字符串中搜索奇数位数：

static const char odd_digits[] = "13579";
std::string number_as_text;
std::cin >> number_as_text;
std::string::size_type  position;
position = number_as_text.find_first_of(odd_digits);
if (position != std::string::npos)
{
  std::cout << "Number has at least one odd digit.\n";
}

从字母列表中搜索任何字符作为字符串。

Answer 2

我做了很少的改动，以保留你原来的思想流。 n=n/10需要在else个案例中执行input as 0。添加了对return(0)案例的处理。当您检测到偶数位时添加了int main() { int n, c; cin >> n; if (n==0) { cout << "Number is zero. So it doesn't have only odd digits" << endl; return(0); } while (n != 0) { c = n % 10; n = n / 10; cout << "n=" << n << " c=" << c << endl; if (c % 2 == 1) { if (n == 0) { cout << "Number has only odd digits" << endl; } } else { cout << "Number doesn't have only odd digits" << endl; return(0); } } return(0); } 语句（继续没有任何意义，因此从那里返回）。

 SELECT * 
 FROM   table_name 
 WHERE  event IN ( 37, 49 ) 
   AND ename NOT IN(SELECT DISTINCT( ename ) 
                    FROM   table_name 
                    WHERE  date IS NOT NULL)

 UNION
 SELECT * 
 FROM   table_name 
 WHERE  event NOT IN ( 37, 49 ) 
   AND ename NOT IN(SELECT DISTINCT( ename ) 
                    FROM   table_name 
                    WHERE  date IS NOT NULL);  

Answer 3

你的问题是这些问题：

else
    cout << " Number doesn't have only odd digits";
    return 0;

编译器将其解释为：

else {
    cout << " Number doesn't have only odd digits";
}
return 0;

这意味着无论return是奇数还是偶数，它总是执行c。要解决此问题，请在else子句中添加显式大括号：

else {
    cout << " Number doesn't have only odd digits";
    return 0;
}

这种错误就是为什么大多数样式指南要求您在if和else之后始终使用大括号，即使只有一行。