我在将整数转换为字符串时遇到问题

Question

namespace C2360_Ch07_Console1_InRange
{
    class InRange
    {

        static void Main(string[] args)
        {
           

            Console.WriteLine("Num:");

            String theLine;
            theLine = Console.ReadLine();

            
            try
            {
                theLine = Convert.ToInt32(Console.ReadLine());//cant convert

            }

            catch (FormatException)
            {
                Console.WriteLine("Input string was not in a correct format.");
            }

            IsWithinRange(theLine);
        }

        static void IsWithinRange(String theLine)
        {
            int Line1;
            Int32.TryParse(Console.ReadLine(), out Line1);

            try
            {

                if ((Line1 < 1) || (Line1 > 10))
                {
                    throw new OverflowException();

                }

            }
            catch (OverflowException)
            {
                Console.WriteLine("number must be between 1 and 10.");
            }

            Console.ReadLine();
        }

    }

}

我想要做的是 convert string 到 integer 以验证 Main 和 IsWithinRange 方法。当我将theLine 转换为Int32.TryParse 时，我收到一个错误，评论所在的位置。这有可能吗？ 有什么帮助吗？

Answer 1

我建议提取一个方法来读取整数值：

   private static int ReadInteger(string title) {
       while (true) {
           Console.WriteLine(title);

           if (int.TryParse(Console.ReadLine(), out int result))
               return result;

           Console.WriteLine("Incorrect syntax. Please, try again");
       }
   }    

然后使用它：

   static void Main(string[] args) {
       int theLine = 0;

       while (true) { 
         theLine = ReadInteger("Num:");

         // theLine is an integer, but we want an extra conditions meet:  
         if (theLine >= 1 && theLine <= 10)
           break;

         Console.WriteLine("The value must be in [1..10] range. Please, try again"); 
       }

       // from now on theLine is an integer in [1..10] range 
       //TODO: put relevant code here
   }

请注意，异常 FormatException、OverflowException 用于异常行为；这里（用户输入验证），好老的if 就够了。

编辑：如果你不想提取方法（为什么？）但保留IsWithinRange，你可以这样写：

   static void Main(string[] args) {
       int theLine = 0;

       while (true) {
           Console.WriteLine("Num:");

           if (!int.TryParse(Console.ReadLine(), out theLine)) {
               Console.WriteLine("Syntax error. Please, try again");

               continue;
           } 

           if (IsWithinRange(theLine))
               break;

           Console.WriteLine("Sorry, the value is out of range. Please, try again"); 
       }
       // from now on theLine is an integer in [1..10] range 
       //TODO: put relevant code here

IsWithinRange 可以在哪里

   // Least Confusion Principle: For "Is [the value] Within Range" question
   // the expected answer is either true or false
   private static bool IsWithinRange(int value) {
       return value >= 1 && value <= 10;
   }