c++ - 为什么这个程序打印两次 "nine"？

Question

现场版:http://cpp.sh/953y6

代码:

#include <iostream>
#include <cstdio>
using namespace std;

int main() {
    // Complete the code.
    int num1 = 8, num2 = 11;
    for(int n = num1; n <= num2; n++){

        if(n <= 9){
            switch(n){
                case 1: cout << "one\n";
                case 2: cout << "two\n";
                case 3: cout << "three\n";
                case 4: cout << "four\n";
                case 5: cout << "five\n";
                case 6: cout << "six\n";
                case 7: cout << "seven\n";
                case 8: cout << "eight\n";
                case 9: cout << "nine\n";
            }

      }
      else if(n % 2 == 0){ //even
            cout << "even\n";
      }
      else if(n > 9 && n %2 == 1){ //odd
            cout << "odd\n";
      }

   }

    return 0;
}

数字 8 到 11 在 for 循环中循环。 if(n <= 9) 应该只被触发两次，当 n 是 8 和当 n 是 9 时。相反，输出是:

eight
nine
nine
even
odd

为什么？

Answer 1

因为你没有 break 并且它是 fall-through case。

case 8: cout << "eight\n";  // <-- need break here
case 9: cout << "nine\n";   // otherwise it's fall-through to here even input is 8

• 好的做法是总是在 switch 的每个 case 之后 break。
• 如果您需要故意失败，请明确说明对失败发表评论。