Question

我最近开始从事Java编程，并认为自己是编程新手。看来我的源代码算法存在问题。我已经验证了所有嵌套的if-else语句，并且它们都符合最后的else语句的算术。它没有正确计算我已经将算法设置为上面的if-else语句。

else语句假设从金额中减去40，然后应用1％的费用。我试过了else语句fee = ((checkAmount - 40) * .01)和fee = ((checkAmount * .01) - 40)

这只是本书的练习

  import java.util.Scanner;
public class ServiceCharge {
public static void main(String[] args)
{
    double checkAmount;
    double fee;
    Scanner kb = new Scanner(System.in);
    System.out.println("I will calulate the service charge to cash your check");
    System.out.print("Enter amount of check: $");
    checkAmount = kb.nextDouble();

    if (checkAmount > 0)
    {
        if (checkAmount <= 10)
        {
            fee = -1;
            System.out.println("$1 service charge");
            checkAmount = checkAmount + fee;
            System.err.println("You have " + checkAmount + " left after service charge.");
        }
        else if ((checkAmount > 10) && (checkAmount <= 100))
        {
            System.out.println("There will be a 10 percent charge.");
            fee = (checkAmount * .10);
            checkAmount = checkAmount - fee;
            System.out.printf("Processing fee: $%.2f\n" , fee);
            System.out.printf("Check amount: $%.2f\n" , checkAmount);
        }
        else if ((checkAmount > 100) && (checkAmount <= 1000))
        {
            System.out.println("There will be a $5 charge plus 5 percent");
            fee = ((checkAmount - 5) * .05);
            checkAmount = (checkAmount - fee);
            System.out.printf("Processing fee: $%.2f\n" , fee);
            System.out.printf("Check amount: $%.2f\n", checkAmount);
        }
        else
        {
            System.out.println("$40 processing fee plus 1 percent");
            fee = ((checkAmount - 40) * .01);
            checkAmount = (checkAmount - fee);
            System.out.printf("Processing fee: $%.2f\n" , fee);
            System.out.printf("Check amount: $%.2f\n" , checkAmount);
        }
        System.out.println("Thanks for using Service Charger." + "\nGood bye");
    }
}
}

Answer 1

对于最后一个else语句，它似乎与其余的陈述有点不同。您使用“hold”存储原始checkAmount值，然后将checkAmount修改为前三个语句的费用。你应该像前一个那样建模最后一个。 checkAmount应为checkAmount = (checkAmount * .01) + 40，然后hold - checkAmount应返回您要查找的值。通过checkAmount = checkAmount - 40，最后一行将返回暂停(checkAmount) - (checkAmount - 40)，这将始终返回40。

Answer 2

            System.out.println("$40 processing fee plus 1 percent");
            fee = ((checkAmount - 40) * .01);

这不是40美元的费用+ 1％。这是一个略低于1％的费用;就像你免费兑现前40美元，然后向其余部分收取1％的费用。

假设1％的费用适用于整个支票，而不是扣除40美元后的剩余费用，则该费用的正确表达是

            fee = 40 + 0.01*checkAmount;

Answer 3

实际上，根据您的原始表达式，您只想收取超过支票金额减去40美元固定费用的1％费用，因此表达式应为：

fee = 40 + (checkAmount - 40) * .01;

您的代码中存在大量重复，这使得更难以查看正在发生的事情，并且如果您决定更改 - 例如 - 要向用户显示的消息，则现在需要更改它4个地点，有一个很大的变化，你忘记在某个地方做，或者你打错了。

良好编程的目标之一是尽可能避免重复。

public class ServiceCharge {
    public static void main(String[] args) {

        Scanner kb = new Scanner(System.in);
        System.out.println("I will calulate the service charge to cash your check");
        System.out.print("Enter amount of check: $");
        double checkAmount = kb.nextDouble();

        if (checkAmount > 0) {
            double fee;
            String feeMessage;
            if (checkAmount <= 10) {
                fee = 1;
                feeMessage = "$1 service charge";
            } else if ((checkAmount > 10) && (checkAmount <= 100)) {
                feeMessage = "10 percent charge.";
                fee = (checkAmount * .10);
            } else if ((checkAmount > 100) && (checkAmount <= 1000)) {
                feeMessage = "$5 charge plus 5 percent";
                fee = 5 + ((checkAmount - 5) * .05);
            } else {
                feeMessage = "$40 processing fee plus 1 percent";
                fee = 40 + ((checkAmount - 40) * .01);
            }
            checkAmount = checkAmount - fee;

            System.out.printf("Fee structure: " + feeMessage);
            System.out.printf("Processing fee: $%.2f\n", fee);
            System.out.printf("Check amount: $%.2f\n", checkAmount);
            System.out.println("Thanks for using Service Charger.\nGood bye");
        }
    }
}

您可能想要研究的下一步是将程序功能分解为函数。例如，您要求金额的部分，计算部分以及显示结果的部分是三个截然不同的部分。您可能希望单独更改这三个中的每一个 - 您可能希望从文件或Web请求中获取输入，并且您可能希望将结果存储在另一个计算中，而不是将其显示给用户。

所以这些可能会分开进行。

然后你可以考虑对象分解 - 也许我们在这里谈论的CheckCashAction对象具有checkAmount，fee，feeStructure和{{的属性1}}。

等

然后好处是你可以分别查看每个步骤并单独测试它们，这样可以更容易地查明错误并维护代码。

初学者Java程序员算术错误

3 个答案: