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有一个我正在玩的小程序。需要确保它检查是否数字，如果不是循环，直到它们是每个输入上的数字并创建一个主要方法和计算器方法有什么帮助吗？

代码在这里 /////////////////////////////////////// //////

        int num1;
        int num2;
        string operand;
        float answer;
        string text1;




        ////enter first number ////
        Console.Write("Please enter a number: ");
        text1 = Console.ReadLine();

        // if number not integer then fail ////
        bool res = int.TryParse(text1, out num1);
        if (!res)
        {
            Console.WriteLine(" FAIL");
            ////enter first number ////
            Console.Write("Please enter a number: ");
            text1 = Console.ReadLine();
        }
        else
        {

        }



        //// enter operand ////
        Console.Write("Please enter an operand (+, -, /, *): ");
        operand = Console.ReadLine();



        // enter second number //
        Console.Write("Please enter the second number: ");
        text1 = Console.ReadLine();

        // if number not integer then fail //
        bool eff = int.TryParse(text1, out num2);
        if (!eff)
            do
            {
                Console.WriteLine(" FAIL");
                // enter second number //
                Console.Write("Please enter the second number: ");
                text1 = Console.ReadLine();
            }
            while (eff == true);
        {

        }
        // converts number to integer ///


        // makes operand answers from each number ////
        switch (operand)
        {
            case "-":
                answer = num1 - num2;
                break;
            case "+":
                answer = num1 + num2;
                break;
            case "/":
                answer = num1 / num2;
                break;
            case "*":
                answer = num1 * num2;
                break;
            default:
                answer = 0;
                break;
        }

        /// converts numbers to string using operand and writes final line ///
        Console.WriteLine(num1.ToString() + " " + operand + " " + num2.ToString() + " =                   
        "+ answer.ToString());
        Console.ReadLine();
        }

        }


        }

        }

        /// converts numbers to string using operand and writes final line ///
        Console.WriteLine(num1.ToString() + " " + operand + " " + num2.ToString() + " = 
        " + answer.ToString());
        Console.ReadLine();
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2 回答 2

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问题似乎与“eff”和“res”的处理有关。如果用户在第一次被询问时键入一个非整数值，那么他们第二次回答什么并不重要，因为 num1 和 num2 没有填充该值。修复它，代码似乎可以工作。

由于输入值似乎在做同样的事情，但提示略有不同，您应该将其移动到一个单独的函数中，如下所示：

    static int GetNumberFromUser(string order)
    {
        string userText = String.Empty;
        int result;

        Console.Write("Please enter {0} number: ", order);
        userText = Console.ReadLine();

        while (!int.TryParse(userText, out result))
        {
            Console.WriteLine("FAILED");
            Console.Write("Please enter {0} number: ", order);
            userText = Console.ReadLine();
        }
        return result;
    }

然后你会打电话给它

    num1 = GetNumberFromUser("first");
    num2 = GetNumberFromUser("second");

此函数负责转换为数字并不断询问，直到用户键入有效值。

“计算器”方法将被复制并将您拥有的开关粘贴到单独的方法中：

    static float Calculator(int num1, string operand, int num2)
    {
        switch (operand)
        {
            case "-":
                return num1 - num2;
            case "+":
                return num1 + num2;
            case "/":
                return num1 / num2;
            case "*":
                return num1 * num2;
            default:
                return 0;
        }
    }

调用使用

        answer = Calculator(num1, operand, num2);

虽然我在做，但结果行很难阅读，我会选择这样的东西

    Console.WriteLine("{0} {1} {2} = {3}", num1, operand, num2, answer);
于 2012-07-11T11:11:08.490 回答
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如果这就是您所需要的并且您正在使用控制台应用程序，您可以使用：

        int num1;
        int num2;
        string operand = string.Empty;
        float answer;
        string text1;

        ////enter first number ////
        Console.Write("Please enter a number: ");
        text1 = Console.ReadLine();

        // if number not integer then fail ////
        bool res = int.TryParse(text1, out num1);
        while (!res)
        {
            Console.WriteLine(" FAIL");
            ////enter first number ////
            Console.Write("Please enter a number: ");
            text1 = Console.ReadLine();
            res = int.TryParse(text1, out num1);
        }

        //// enter operand ////
        while (operand == string.Empty || operand.Length > 1 || !(new char[] { '+', '-', '*', '/' }).Contains(char.Parse(operand)))
        {
            Console.Write("Please enter an operand (+, -, /, *): ");
            operand = Console.ReadLine();
        }


        // enter second number //
        Console.Write("Please enter the second number: ");
        text1 = Console.ReadLine();

        // if number not integer then fail //
        bool eff = int.TryParse(text1, out num2);
        while (!eff)
        {
            Console.WriteLine(" FAIL");
            // enter second number //
            Console.Write("Please enter the second number: ");
            text1 = Console.ReadLine();
            eff = int.TryParse(text1, out num2);

        }
        // converts number to integer ///


        // makes operand answers from each number ////
        switch (operand)
        {
            case "-":
                answer = num1 - num2;
                break;
            case "+":
                answer = num1 + num2;
                break;
            case "/":
                if (num2 == 0)
                {
                    Console.WriteLine("Divide By Zero Error");
                    return;
                }
                answer = num1 / num2;
                break;
            case "*":
                answer = num1 * num2;
                break;
            default:
                answer = 0;
                break;
        }

        /// converts numbers to string using operand and writes final line ///
        Console.WriteLine(num1.ToString() + " " + operand + " " + num2.ToString() + " =  "+ answer.ToString());
        Console.ReadLine();

为什么你需要为这么简单的事情创建一个单独的方法。我发现这个问题太简单了，所以只问这是否是必需的？你本可以再努力一点，然后自己写这个。这没有什么微不足道的。我只是假设您是编程新手。

于 2012-07-11T11:10:41.787 回答