初学者Java程序员算术错误 我最近开始从事java编程，并把自己当作编程新手。看来我的源代码算法有问题。我已经验证了所有嵌套的if-else语句，它们都符合最终else语句的算法。它计算不正确，我已经按照上面的if-else语句设置了算法

else语句假定从金额中减去40，然后应用1%的费用。我已经尝试了else语句

fee=（（checkAmount-40）*.01）

和

fee=（（checkAmount*.01）-40）

这只是本书中的一个练习

  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");
    }
}
}

import java.util.Scanner；
公共课服务费{
公共静态void main（字符串[]args）
{
双重支票金额；
双倍收费；
扫描仪kb=新扫描仪（System.in）；
System.out.println（“我将计算服务费以兑现您的支票”）；
System.out.print（“输入支票金额：$”；
checkAmount=kb.nextDouble（）；
如果（支票金额>0）
{
如果（支票金额10）和（支票金额100）和（支票金额
这不是40美元+1%的费用。这是略低于1%的费用；这就像你免费兑现前40美元，然后对其余部分收取1%的费用

假设1%的费用适用于整张支票，而不是减去40美元后剩下的钱，那么正确的费用表达式是

            fee = 40 + 0.01*checkAmount;

实际上，根据您最初的表达，您只想收取支票金额的1%费用减去40美元的固定费用，因此表达应该是：

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

您的代码中存在大量重复，这使得查看发生了什么变得更加困难，如果您决定更改—例如—您想向用户显示的消息，您现在需要在4个位置进行更改，并且有一个很大的更改，您忘记在某处执行，或者您输入了一个错误

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");
        }
    }
}

公共类服务费{
公共静态void main（字符串[]args）{
扫描仪kb=新扫描仪（System.in）；
System.out.println（“我将计算服务费以兑现您的支票”）；
System.out.print（“输入支票金额：$”；
double checkAmount=kb.nextDouble（）；
如果（支票金额>0）{
双倍收费；
字符串消息；
如果（checkAmount 10）&（checkAmount 100）&&（checkAmount用于最后一个else语句，它似乎与您的其余语句有点不一致。您使用的是“hold”要存储原始的
checkAmount
值，然后修改
checkAmount
作为前三条语句的费用。您应该将最后一条语句建模为前一条语句。
checkAmount
应该是
checkAmount=（checkAmount*.01）+40
，则
hold-checkAmount
应返回您正在查找的值。通过使用
checkAmount=checkAmount-40
，最后一行返回hold
（checkAmount）-（checkAmount-40）
，它将始终返回
40
实际发生了什么？出现了什么明显的错误？您是否收到错误消息？如果收到，请将其显示给我们。您是否收到错误结果？如果收到，请将其显示给我们，并向我们显示应该发生的情况。您离浮点错误还有多远？您知道吗？40加1%的费用是
费用=40+支票金额*0.01；checkAmount-=fee；
，对吗？虽然你可能是编程新手，但我认为你在学校学过一些数学。仅仅因为你在学校学过，并不意味着它也不适用于编程。谢谢你的澄清。
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");
        }
    }
}