C# C添加方法并检查是否添加了数字

有一个我正在玩的小程序。需要确保它检查数字，如果不是循环，直到它们在每个输入上都是一个数字，并创建一个主方法和计算器方法有帮助吗

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();

---

如果这是您所需的全部，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();

---

为什么要为这么简单的东西创建一个单独的方法。我发现这个问题太简单了，所以我只是想问一下这是否是必要的？你本可以再努力一点，自己写这个。这里面没有什么琐碎的东西。我只是假设你是编程新手

问题似乎在于eff和res的处理。如果用户在第一次被询问时键入一个非整数值，第二次回答什么并不重要，因为num1和num2没有填充该值。解决了这个问题，代码就可以正常工作了

由于输入值似乎是在做同样的事情，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;
    }

然后你可以通过打电话来称呼它

    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

只需复制计算器方法并将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

        answer = Calculator(num1, operand, num2);

当我在做这件事的时候，结果是很难理解的，我会选择这样的东西

    Console.WriteLine("{0} {1} {2} = {3}", num1, operand, num2, answer);