在循環中檢查和使用兩個數組

Question

有什么方法可以簡化以下代碼。 粗略設置的方式是掃描值，但是如果輸入引發異常，則需要說出nonono並重​​新搜索該值。 我需要像這樣收集x和y值，然后才能以科學的計算器方式對其進行操作。 輸入字符串“ RESULT” =先前計算的答案是必要條件。 這兩個要求x和y的循環是如此相似，只有“第一個操作數”和“ x =答案”，而“第二個操作數”和“ y =答案”不同。 那么，有什么方法可以優化此代碼，以便僅需一個循環，因為兩者是如此相似？ 這是代碼。

    String operand;
    double x = 0;
    double y = 0;

    //These two arrays are the differences between both of the loops that follow. Everything besides first, x and second, y are the same
    String arr[] = {"first", "second"};
    Double var[] = {x, y};

    boolean operandLoop1 = false;
    //x
    while (!operandLoop1) {
        System.out.print("Enter " + arr[0] + " operand: ");
        operand = calcOption.next(); // retrieve first value
        if (operand.equals("RESULT")) {
            var[0] = answer; // If I want to use the previous result as my input
            operandLoop1 = true;
        } else {
            try {
                var[0] = Double.parseDouble(operand); // Assumes that if it isn't RESULT, then I'll want to put in a number
                operandLoop1 = true;
            } catch (NumberFormatException nfe) { // necessary if I type anything else in besides RESULT and a double
                System.out.print("Error: Invalid input! Correct inputs are any real number and \"RESULT\"!");
            }
        }
    }

    boolean operandLoop2 = false;
    //y
    while (!operandLoop2) {
        System.out.print("Enter" + arr[1] + " operand: ");
        operand = calcOption.next(); // retrieve second value
        if (operand.equals("RESULT")) {
            var[1] = answer; // If I want to use the previous result as my input
            operandLoop2 = true;
        } else {
            try {
                var[1] = Double.parseDouble(operand); // Assumes that if it isn't RESULT, then I'll want to put in a number
                operandLoop2 = true;
            } catch (NumberFormatException nfe) { // necessary if I type anything else in besides RESULT and a double
                System.out.print("Error: Invalid input! Correct inputs are any real number and \"RESULT\"!");
            }
        }
    }

關於長度，我很抱歉，但是希望我能得到大約一半的長度。

Answer 1

由於這兩個部分之間唯一的區別是數組索引，因此可以使用for循環，如下所示：

for (int i = 0; i < 2; i++) {
    boolean operandLoop = false;
    while (!operandLoop) {
        System.out.print("Enter " + arr[i] + " operand: ");
        operand = calcOption.next(); // retrieve value
        if (operand.equals("RESULT")) {
            var[i] = answer; // If I want to use the previous result as my input
            operandLoop = true;
        } else {
            try {
                var[i] = Double.parseDouble(operand); // Assumes that if it isn't RESULT, then I'll want to put in a number
                operandLoop = true;
            } catch (NumberFormatException nfe) { // necessary if I type anything else in besides RESULT and a double
                System.out.print("Error: Invalid input! Correct inputs are any real number and \"RESULT\"!");
            }
        }
    }
}

您也可以使其成為一種方法，傳入calcOption ， answer和序數（ arr[0] ）的參數，並用return語句替換var[0]所有賦值。 我不知道calcOption的類型，但是看起來像這樣：

Double methodName(Object calcOption, Double answer, String ordinal) {
    boolean operandLoop = false;
    while (!operandLoop) {
        System.out.print("Enter " + ordinal + " operand: ");
        String operand = calcOption.next(); // retrieve value
        if (operand.equals("RESULT")) {
            return answer; // If I want to use the previous result as my input
        } else {
            try {
                return Double.parseDouble(operand); // Assumes that if it isn't RESULT, then I'll want to put in a number
            } catch (NumberFormatException nfe) { // necessary if I type anything else in besides RESULT and a double
                throw new RuntimeException("Error: Invalid input! Correct inputs are any real number and \"RESULT\"!");
            }
        }
    }
}