java - 如何在程序得到有效答案之前提示用户，怎么做N次？

Question

我正在尝试制作一个简单的程序，如果用户在数组中输入一个数字，则该数组数字等于的数字将会出现。它工作完美，除非用户输入一个不在数组中的数字，然后在数组中输入一个数字。发生的事情是程序没有在正确的地方显示答案。

Scanner ethan = new Scanner(System.in);

System.out.println("Enter array place: ");
int amount = ethan.nextInt();

int array[] = {2, 4, 9};

for(int i = 0; i<amount; i++)
{
    if (amount == 0) {
        System.out.println(array[0]);
    } else if (amount == 1) {
        System.out.println(array[1]);
    } else if (amount == 2) {
        System.out.println(array[2]);
    } else {
        System.out.println("Enter integer in array:");
        amount = ethan.nextInt();
    }
}

Answer 1

你不应该在 for 循环中使用输入的数字

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

如果我输入 0 会怎样？该程序不会运行单个循环。

试试这个，它应该是有效的。我在代码中为您做了一些评论。

int array[] = {2, 4, 9};
Scanner ethan = new Scanner(System.in);

/**
 * This gets one valid number from the user
 */
public void getOneNumber() {
    /**
     * This is our "switch" variable
     */
    boolean validIndexGiven = false;

    while (!validIndexGiven)

    {
        /** This is actually called "array index"*/
        System.out.println("Enter array index: 0 to " + (array.length - 1));
        int index = ethan.nextInt();

        /** Here we check if given integer is in the bounds of array.
         * No need to check every possible value.
         * What if we had array[100]? */
        if (index >= 0 && index < array.length) {
            /** Amount was a valid number. */
            System.out.println(array[index]);
            /** Toggling the switch */
            validIndexGiven = true;
        }
    }
}

/**
 * This prompts the user for numbers as many times as there are elements in array
 */
public void getManyNumbers() {
    int length = array.length;
    for (int i = 0; i < length; i++) {
        getOneNumber();
    }
}