Question

我正在制作一个简单的计算器，但是在我能够做到这一点之前，我需要制作一个算法来将用户输入转换为一致的格式。用户可以输入数字和运算符以及括号。该程序处理数字和运算符没有问题，但由于某些未知原因，只要循环遇到括号，它就会抛出错误。

我一直在尝试调试过去几个小时的代码，但我似乎无法弄清楚为什么会发生这种情况？

/** These are the possible operators */
private static final String OPERATORS = "+-/*%^()[]{}";

/** This is an ArrayList of all the discrete
    things (operators/operands) making up an input.
    This is really just getting rid of the spaces,
    and dividing up the "stuff" into manageable pieces.
*/
static ArrayList<String> input = new ArrayList<String>();

public static ArrayList inputCleaner(String postfix) {
    StringBuilder poop = new StringBuilder();
    String doody = postfix.replace(" ", "");
    try {
        for (int i = 0; i < doody.length(); i++) {
            char c = doody.charAt(i);
            boolean isNum = (c >= '0' && c <= '9');
            if (isNum) {
                poop.append(c);
                if (i == doody.length() - 1) {
                    input.add(poop.toString());
                    poop.delete(0, poop.length());
                }
            }
            else if (c == '.') {
                for (int j = 0; j < poop.length(); j++) {
                    if (poop.charAt(j) == '.') {
                        throw new SyntaxErrorException("You can't have two decimals in a number.");
                    }
                    else if (j == poop.length() - 1) {
                        poop.append(c);
                    }
                } 
                if (i == doody.length() - 1) {
                    throw new SyntaxErrorException("You can't end your equation with a decimal!");
                }
            }
            else if (OPERATORS.indexOf(c) != -1 && poop.length() != 0) {
                input.add(poop.toString());
                poop.delete(0, poop.length());
                poop.append(c);
                input.add(poop.toString());
                poop.delete(0, poop.length());  
            }
            else {
                throw new SyntaxErrorException("Make sure your input only contains numbers, operators, or parantheses/brackets/braces.");
            }
        }
        return input;
    }
    catch (SyntaxErrorException exc) {
        System.out.println("That didn't work, something was wrong with the syntax.");
        return input;
    }
}

public static void main(String[] args) {
    ArrayList test = new ArrayList();
    Scanner f = new Scanner(System.in);
    System.out.println("Please insert an argument: \n");
    String g = f.nextLine();
    test = inputCleaner(g);

    for (int z = 0; z < test.size(); z++) {
        System.out.println(test.get(z));
    }
}

Answer 1

我知道我的答案不是很好因为我对它没有很好的解释（由于工作中的大脑过热）但这种变化不会像你的情况那样抛出异常（可能是你的答案;））< / p>

我没有在条件中使用poop.length() != 0，而只是将其更改为poop != null而瞧...现在没有Exceptions。评论你是否可以解释我这个。

当用户输入括号时，Java程序会变得怪异

1 个答案: