Question

我必须用Java创建一个程序，该程序使用一种方法来计算未来的投资价值，用户可以投入投资金额和利率。它必须显示在表格中并具有年份1-30。

我有一个错误，我无法弄清楚如何修复。我//Title main中的错误位于futureInvestmentValue下。错误告诉我

futureInvestmentValue无法解析为变量

这是我的代码：

import java.util.Scanner;
public class futureInvestmentValue {    
    public static double futureInvestmentValue(double investmentAmount, 
        double monthlyInterestRate, int years){
        double futureInvestmentValue = 1;

        for (years = 1; years <= 30; years++) {
            monthlyInterestRate = years/1200;
            futureInvestmentValue = 
                investmentAmount * Math.pow((1+monthlyInterestRate), years*12);
        }
        return futureInvestmentValue;
    }           

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        //Get investment amount and yearly interest rate from user
        Scanner input = new Scanner (System.in);
        System.out.print("Enter investment amount, for example 100:" + 
            "Enter yearly interest rate, for example 5.25:\n");
        double investmentAmount = input.nextDouble();
        double annualInterestRate = input.nextDouble();

        //Title
        System.out.print("Years\t" + "\t Future Value");

        for (int years = 1; years <= 30; years++) {
            System.out.print(years);
            System.out.printf("%4d", futureInvestmentValue);

        }
        System.out.println();               
    }    
}

Answer 1

在main方法中，您引用的是变量而不是函数名...

System.out.printf("%4d", futureInvestmentValue);

以上应该是：

System.out.printf("%4d", futureInvestmentValue(investmentAmount, ..... ));

顺便说一句，这就是为变量和函数赋予相同的名称。不要这样做。

错误：futureInvestmentValue无法解析为变量

1 个答案: