Question

我正在建立一个计算和输出以下值的程序：5年后1％的利率，价值范围从$ 1000- $ 20000（1000,2000,3000等），没有复合。

Ex）1000为1050，2000为2100（公式为A = P（1 + rt））。

目前，我试图以这样的方式编写我的代码：运行时，它会在垂直的不同行上列出1000 - 20000的所有值，然后在没有复合的兴趣之后列出所有值：

1000 2000 ... 20000 1050.0 2100.0 ... 21000.0

（上述数字意味着垂直显示在另一个之下）

我尝试自学的一个项目是将一个值从一个方法返回到main方法。在这种情况下，我的calculateAmountNoCompounding方法有一个返回值，我设置为`double testValue =（j *（1 + .01 * NUMBER_OF_YEARS））;.然而，当我尝试通过将它放在打印行函数中将其返回到main方法时，我得到错误“.class expected”。我也在第25行的实际return语句中得到了同样的错误。是否有一种简单有效的方法来处理这个问题？

public class DepositCalculator {

public static final double ANNUAL_INTEREST_RATE = 1.0;
public static final int NUMBER_OF_YEARS = 5;
public static final int MONTHS_PER_YEAR = 12;
public static final int LOW_DEPOSIT_AMOUNT = 1000;
public static final int HIGH_DEPOSIT_AMOUNT = 20000;
public static final int DEPOSIT_AMOUNT_INCREMENT = 1000;

public static void main(String[] args) {

    for (int i = 0; i <= HIGH_DEPOSIT_AMOUNT - DEPOSIT_AMOUNT_INCREMENT; i = i + DEPOSIT_AMOUNT_INCREMENT) {
        System.out.println(LOW_DEPOSIT_AMOUNT + i);

    }     
    System.out.println(double testValue);

}

public static double calculateAmountNoCompounding(double deposit, double annualInterestRate, int numberOfYears) {

    for (double j = LOW_DEPOSIT_AMOUNT; j <= HIGH_DEPOSIT_AMOUNT - DEPOSIT_AMOUNT_INCREMENT; j = j + DEPOSIT_AMOUNT_INCREMENT) {
        double testValue = (j * (1 + .01 * NUMBER_OF_YEARS));
    }
    return double testValue;
}

}

Answer 1

您无法在calculateAmountNoCompounding中致电main。
double此处是一个类，因此请将其写为Double。
您应该在Double testValue;循环之上初始化for，以便在您想要返回时显示它。
应该在不指定类型的情况下调用现有变量。你不能这样做return double testValue;，只做return testValue;。由于3.，Java已经知道testValue是double。这同样适用于System.out.println(double testValue);。

Answer 2

它应该是return testValue;而不是return double testValue。您还需要在for循环

之外声明变量testValue

Answer 3

在打印或返回语句之前，您不需要使用“double”。

返回应该只是testValue并打印返回的值：

double answer = calculateAmountNoCompounding(...);
System.out.println(answer);

或者可以直接打印

System.out.println(calculateAmountNoCompounding(...));

Answer 4

好的，所以我真的不想给你所有这些答案。但是我已经使用了你的代码并在那里放了一个例子

public class DepositCalculator {

public static final double ANNUAL_INTEREST_RATE = 1.0;
public static final int NUMBER_OF_YEARS = 5;
public static final int MONTHS_PER_YEAR = 12;
public static final int LOW_DEPOSIT_AMOUNT = 1000;
public static final int HIGH_DEPOSIT_AMOUNT = 20000;
public static final int DEPOSIT_AMOUNT_INCREMENT = 1000;

public static void main(String[] args) {

    double result = 0.0;
    double deposit = 0.0;
    double annualInterestRate = 0.0;
    int numberOfYears = 0;

    //  you need to think about this
    /*for (int i = 0; i <= HIGH_DEPOSIT_AMOUNT - DEPOSIT_AMOUNT_INCREMENT; i = i + DEPOSIT_AMOUNT_INCREMENT) {
      //  System.out.println(LOW_DEPOSIT_AMOUNT + i);

    }*/    

    //  you need to set the deposit amount, annualinterestrate and number of years

    //result = calculateAmountNoCompounding(deposit, annualInterestRate, numberOfYears);

    //  here is an example deposit is 5000, annual interest is 10 and numberofyears is 5
    result = calculateAmountNoCompounding(5000, 10, 5);



    System.out.println("result is " +result);
}

public static double calculateAmountNoCompounding(double deposit, double annualInterestRate, int numberOfYears) {
    double testValue = 0.0;
    //  you need to think about this
    //for (double j = LOW_DEPOSIT_AMOUNT; j <= HIGH_DEPOSIT_AMOUNT - DEPOSIT_AMOUNT_INCREMENT; j = j + DEPOSIT_AMOUNT_INCREMENT) {
        testValue = (deposit * (1 + .01 * numberOfYears));
    //}
    return testValue;
}

}

我希望有所帮助。

将值返回到main方法

4 个答案: