Question

有什么方法可以简化以下代码。粗略设置的方式是扫描值，但是如果输入引发异常，则需要说出nonono并重新搜索该值。我需要像这样收集x和y值，然后才能以科学的计算器方式对其进行操作。输入字符串“ RESULT” =先前计算的答案是必要条件。这两个要求x和y的循环是如此相似，只有“第一个操作数”和“ x =答案”，而“第二个操作数”和“ y =答案”不同。那么，有什么方法可以优化此代码，以便仅需一个循环，因为两者是如此相似？这是代码。

    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\"!");
            }
        }
    }

很抱歉，但希望我能将其长度缩短一半。

Answer 1

由于这两个部分之间唯一的区别是数组索引，因此可以使用如下所示的for循环：

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\"!");
            }
        }
    }
}

您还可以使其成为一种方法，传入calcOption，answer和序数（arr[0]）的参数，并替换{{1}的所有赋值}和return语句。我不知道var[0]的类型，但是看起来像这样：

calcOption

在循环中检查和使用两个数组

1 个答案: