输入整数时如何修复打印输出

Question

如果我输入“0”、“1”、“2”或“3”（按特定顺序），程序的输出是完美的。如果我以不同的顺序输入整数，代码将无法正常工作。例如，如果我首先选择“2”，我的代码希望我输入整数总共 3 次，以便给我正确的输出。有人可以告诉我我做错了什么吗？

我尝试过使用 else-if 语句，当我输入除 '0' 以外的任何内容时，我需要输入整数的总次数等于索引号。例如，如果我输入'2'，我必须总共输入3次才能得到我想要的输出。

System.out.println("Please input a number between zero to 3");

            for (int i = 0; i < 4; i++) {

                if (sc.nextInt() == 0) {
                    System.out.println("You have selected " + right);
                }
                if (sc.nextInt() == 1) {
                    System.out.println("You have selected " + left);
                }
                if (sc.nextInt() == 2) {
                    System.out.println("You have selected " + up);
                }
                if (sc.nextInt() == 3) {
                    System.out.println("You have selected " + down);
                    break;
                }
            }

我的预期输出应该是：

This program simulates the 4 arrows RIGHT, LEFT, UP, DOWN using the numbers 0, 1, 2, 3 respectively
Please input a number between zero to 3
3
You have selected DOWN
1
You have selected LEFT
0
You have selected RIGHT
2
You have selected UP

Process finished with exit code 0

---

当我将它们按正确的顺序放置时，就会发生此输出。如果我从输入“1”开始，就会发生这种情况：

This program simulates the 4 arrows RIGHT, LEFT, UP, DOWN using the numbers 0, 1, 2, 3 respectively
Please input a number between zero to 3
1
1
You have selected LEFT

Answer 1

将您的逻辑更改为：

for (int i = 0; i < 4; i++) {
    System.out.println("Please input a number between zero to 3");
    // use input and don't advance the scanner every time
    int input = sc.nextInt();

    if (input == 0) {
        System.out.println("You have selected " + right);
    }
    if (input == 1) {
        System.out.println("You have selected " + left);
    }

    // so on and so forth

}

通过使用sc.nextInt() 四次 次，您正在寻找不存在的输入的下一个标记。因此，为 for 循环的 每个 运行获取输入，它将按预期工作。