好.開始了.我已經(jīng)修改了你的程序了一點點.以下是修改程序時出現(xiàn)的問題.

1.任何類中都沒有“writeOutput()方法.因此我不得不刪除那段代碼.

2.必須引入while循環(huán)才能使其成為迭代過程.

3.在讀取運算符字符串后向下移動operator.charAt(0)方法.

由于您使用的是switch語句,因此無法將運算符值與“nCr”字符串值進行比較.要使用此方法,您必須使用equals()方法使用字符串比較.

最后但并非最不重要的是,操作的結(jié)果應該是下一輪計算的輸入之一.意味著第一個操作的結(jié)果將被傳遞到下一個操作的第一個參數(shù).

import java.util.Scanner;

public class RPNCalc

{

public static void main(String[]args)

{

Scanner keyboard = new Scanner(System.in);

double v1, v2;

// String operator = keyboard.nextLine();

// char symbol = operator.charAt(0);

char operator = ' ';//First time around, set this to an something other than 'q'

String operatorString = " ";

System.out.print("Enter a value v1: ");

v1 = keyboard.nextDouble();

System.out.println();

while(operator != 'q')

{

System.out.print("Enter a value v2: ");

v2 = keyboard.nextDouble();

System.out.println();

System.out.print("Enter one of the valid operators +, -, *, /, nCr,: ");

operatorString = keyboard.next();//nextLine() doesn't wait until the user hit enter

operator = operatorString.charAt(0);

switch (operator)

{

case'+':

v1 = sum(v1, v2);

System.out.println(v1);

break;

case'-':

v1 = minus(v1, v2);

System.out.println(v1);

break;

case'*':

v1 = times(v1, v2);

System.out.println(v1);

break;

case'/':

v1 = divide(v1, v2);

System.out.println(v1);

break;

case'q':

System.out.println("Your last result was " );

default:

System.out.println("You must choose an appropriate operator .");

}

}

}

double value1, value2;

int n;

public static double sum(double value1,double value2)

{

double newSum = value1 + value2;

return newSum;

}

public static double minus(double value1, double value2)

{

double newMinus = value1 - value2;

return newMinus;

}

public static double times(double value1, double value2)

{

double newTimes = value1 * value2;

return newTimes;

}

public static double divide(double value1, double value2)

{

if (value2 == 0)

{

System.out.println("Division by Zero. Try again");

return value1;

}

double newDivide = value1 / value2;

return newDivide;

}

}

總結(jié)

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