繁体   English   中英

简单计算器操作

[英]Simple Calculator Operation

我试图简化我的计算器程序的长代码,但我有一个障碍。 我为每个计算器操作员都有一个新的if语句,但我想要做的是允许用户在一行上手动输入他们想要执行的整个操作并让代码计算它。

这就是我所拥有的:

do {
        System.out.println("What function would you like to perform?");
        System.out.print("Exit Calculator (Q), Add (+), Subtract (-), Multiply (x), Divide (/): ");
        maininput = in.next();

        if (maininput.equals("+")) {
            System.out.print("Enter the first number to add: ");
            num1 = in.nextDouble();
            System.out.print("Enter the second number to add: ");
            num2 = in.nextDouble();
            System.out.println();

            answer = num1 + num2;

            System.out.println(num1 + " + " + num2 + " = " + answer);
            System.out.println();
        }
        else if (maininput.equals("-")) {
            System.out.print("Enter the first number to subtract: ");
            num1 = in.nextDouble();
            System.out.print("Enter the second number to subtract: ");
            num2 = in.nextDouble();
            System.out.println();

            answer = num1 - num2;

            System.out.println(num1 + " - " + num2 + " = " + answer);
            System.out.println();
        }
        else if(maininput.equals("x")) {
            System.out.print("Enter the first number to multiply: ");
            num1 = in.nextDouble();
            System.out.print("Enter the second number to multiply: ");
            num2 = in.nextDouble();
            System.out.println();

            answer = num1 * num2;

            System.out.println(num1 + " x " + num2 + " = " + answer);
            System.out.println();
        }
        else if(maininput.equals("/")) {
            System.out.print("Enter the first number to divide: ");
            num1 = in.nextDouble();
            do {
                System.out.print("Enter the second number to divide: ");
                num2 = in.nextDouble();
                System.out.println();
                if (num2 == 0) {
                    System.out.println("Cannot divide by 0!  Please enter a different number.");
                }
            } while (num2 == 0);

            answer = num1 / num2;

            System.out.println(num1 + " / " + num2 + " = " + answer);
            System.out.println();
        }
        else if(maininput.equals("Q") || maininput.equals("q") || maininput.equals("EXIT") || maininput.equals("exit")) {
            in.close();
            System.exit(0);
        }
        else {
            System.out.println(maininput + " is not a valid operand.  Please try again.");
            System.out.println();
        }
    } while (maininput != "Q" && maininput != "q");

这就是我想要的输出:

Enter operation:
4 * 6
4 * 6 = 24

应该可以在这一行输入任何操作。 我不是要求你为我写我的计算器,我问的是如何允许计算机读取一行的整个操作并计算它,然后打印它。

如果您使用扫描仪readLine,那么您可以读取整行

例如

 4 * 6

然后可以拆分该行以获得三个令牌

 String tokens [] = line.split (" ");

然后你可以看到基于令牌[1]的操作

 if (token[1].equals ("-") {

      //lets minus token[2] from token[0]
      // need to convert String to Number
 }

您可以使用String.split并将其存储在数组中。 然后它将返回一个字符串数组,将它们解析回整数。 做你想要的操作。 x变量将是结果。

    if(maininput.contains("+")) {
        String[] stringarr = string.split("\\+");
        int x = Integer.parseInt(stringarr[0])  +  Integer.parseInt(stringarr[1]);
        System.out.println(stringarr[0] + " + " + stringarr[1] + " = " + x);
    } else if(maininput.contains("-")) {
        String[] stringarr = string.split("\\-");
        int x = Integer.parseInt(stringarr[0])  -  Integer.parseInt(stringarr[1]);
        System.out.println(stringarr[0] + " - " + stringarr[1] + " = " x);
    } 

... 等等。

您可以尝试使用Pattern对象解析该行,如下所示:

Pattern opPattern = Pattern.compile("(\\d+) *([+-*/]) *(\\d+)");
Matcher matcher = opPattern.matcher(userLine);
if(matcher.find()) {
    int op1 = Integer.toValue(matcher.group(1));
    int op2 = Integer.toValue(matcher.group(3));
    String op = matcher.group(2);

    if(op.equals("+")) {
        // do + op ...
    } else ... {
        // etc...
    }

} else {
    // error in line, not the form of an operation
}

看看javadoc,因为我不确定我是否使用了正确的方法名称等,只是试图说明这个想法......

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM