最近,我开始了Java编程,并认为自己是编程新手。我的源代码算法似乎有点问题。我已经验证了所有嵌套的if- all语句,它们都达到了最后一个the语句的算术。它没有正确计算,我已经设置了算法,就像上面的if- set语句一样。
then语句应该从金额中减去40,然后应用1%的收费。我尝试过使用fee = ((checkAmount - 40) * .01)和fee = ((checkAmount * .01) - 40)语句
这只是一本书上的练习
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");
}
}
}发布于 2014-02-08 05:56:40
System.out.println("$40 processing fee plus 1 percent");
fee = ((checkAmount - 40) * .01);这可不是40美元加1%的费用。这是一个略低于1%的费用,就好像你免费兑现了前40美元,然后对剩下的部分收取了1%的费用。
假设1%的费用适用于整个支票,而不是减去40美元后剩下的,那么正确的表示是
fee = 40 + 0.01*checkAmount;发布于 2014-02-08 04:52:24
对于最后一条of语句,它似乎与您的其余语句有些出入。您使用“持有”来存储原始的checkAmount值,然后修改checkAmount作为前三条语句的费用。你应该把最后一个模仿成它之前的那个。checkAmount应该是checkAmount = (checkAmount * .01) + 40,然后hold - checkAmount应该返回您想要的值。通过使用checkAmount = checkAmount - 40,最后一行将返回Hold(checkAmount) - (checkAmount - 40),这将始终返回40。
发布于 2014-02-08 06:40:07
其实,你只想按你原来的说法,只收取支票金额的1%,减去40元的固定费用,所以应该是:
fee = 40 + (checkAmount - 40) * .01;您的代码中有很多重复,这使得您很难看到正在发生的事情,如果您决定更改--例如,您想向用户显示的信息--您现在需要在4个位置更改它,并且有一个很大的变化,您忘记在某个地方做它,或者您做了一个错误。
好的方案编制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");
}
}
}下一步您可能需要研究的是将程序从功能上分解为函数。例如,您要求金额的部分,您进行计算的部分,以及显示结果的部分,是三个非常不同的部分。这三个结果中的每一个都可能需要单独更改--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,而不是显示给用户。
所以它们可以分成不同的功能。
然后您可以考虑对象分解--也许我们在这里讨论的是一个CheckCashAction对象,它具有checkAmount、fee、feeStructure和payoutAmount的属性。
等。
这样做的好处是,您可以分别查看每个步骤,并隔离地测试它们,这使得查找bug和维护代码变得更加容易。
https://stackoverflow.com/questions/21642489
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