我需要从我在main方法中调用的方法打印出应税收入。但是方法中的循环只返回一个值

Question

这是我的代码，计算税法中的循环只会返回第一个nmber甚至不正确。我怎样才能返回正确的值。在每个循环中，必须选择一种类型的税级，您输入的数字为1-4。这也将转到方法循环。

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

        Scanner scan = new Scanner(System.in);
        System.out.println(" Enter in status\n (1 for single\n,2 for married joint, 3 for Married         separate, 4 for Head of House: ");
        int status = scan.nextInt();
        //if status to print out status name
        if (status == 1)
            System.out.println ("Taxable\t single\nIncome");
        else if (status == 2)
            System.out.println ("Taxable\tMarried Jointly\nQualified Widow\nIncome");
        else if (status == 3)
            System.out.println ("Taxable\tMarried separate\nIncome");
        else if (status == 4)
            System.out.println ("Taxable\tHead of House\nIncome");
        int taxableIncome = 50000;


        for (int i = 50000; i <= 60000; i += 50)
            System.out.println (i + "\t" + Math.round(computeTax(status, taxableIncome)));

    }

    public static double computeTax (int status, double taxableIncome) {
        double tax=0;
        for (int k = 50000; k <= 60000; k =+ 50) {
            if (status == 1)
                taxableIncome = 4675 + ((k-33950)*.25);
            else if(status == 2)
                taxableIncome = 16701 + (k-16701)*.15;
            else if(status == 3)
                taxableIncome = 4675 + (k-33951)*.25;
            else if(status == 4)
                taxableIncome = 6227.5 + (k-33549)*.25;
            return taxableIncome[k];
        }
    }
}

Answer 1

To＆＃34; return＆＃34;不使用数组的多个值，在方法中设置全局变量，然后返回第二个值。这是一些概念代码。

public static double returnVal2;

public static void method(){
    double returnVal1 = otherMethod();
    // returnVal1 now holds 1 return value, and returnVal2 now holds the other.
}

public static double otherMethod(){
    returnVal2 = 0; // Change 0 to second value that you want to return.
    return 0;       // Change 0 to the first value that you want to return.
}