从控制台获取c#输入但不返回正确的输出[重复]

Question

C# 程序试图从用户那里获取输入并使用这两个数字来查找产品。

using System;

namespace Program
{
    class MainClass
    {
        static int computeProd1(int num1, int num2)
        {
            return (num1 * num2);
        }

        public static void Main (string[] args)
        {
            int first, second;

            Console.Write ("enter first number: ");
            first = Convert.ToInt32 (Console.ReadLine ());
            Console.Write ("enter 2nd number: ");
            second = Convert.ToInt32 (Console.ReadLine ());

            int product = computeProd1 (first, second);
            //Console.WriteLine ("\n First number is: {0}" , first);
            //Console.WriteLine ("\n Second number is: {0}" , second);
            Console.WriteLine ("\nTheir product is: {0}" , product);
            Console.ReadKey ();     
        }
    }
}

无论我输入什么，第一个和第二个数字都已经输出了一个数字。

Answer 1

console.Read 返回输入的字符的 ascii 码而不是 int 值

试试这个代码，它会要求输入数字（只有数字），然后返回产品：

class Program
{
    static int computeProd1(int num1, int num2)
    {
        return (num1 * num2);

    }

    public static void Main(string[] args)
    {

        int? first = null, second = null;
        int first_i = 0, second_i = 0;

        while (!first.HasValue)
        {
            Console.WriteLine("enter first number");

            if (int.TryParse(Console.ReadLine(), out first_i))
            {
                first = first_i;
            }
        }

        while (!second.HasValue)
        {
            Console.WriteLine("enter second number");

            if (int.TryParse(Console.ReadLine(), out second_i))
            {
                second = first_i;
            }
        }

        int product = computeProd1(first_i, second_i);
        Console.WriteLine("Their product is:\t" + product);
        Console.Read();
    }
}

Answer 2

也许试试这个代码。

更新

使用 int.tryParse() 的更多失败证明方法。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApplication1
{
    class Program
    {
        static int computeProd1(int num1, int num2)
        {
            return (num1 * num2);

        }

        public static void Main(string[] args)

        {
            try
            {
                int first, second, product = 0;

                if (int.TryParse(Console.ReadLine(), out first))
                {
                    if (int.TryParse(Console.ReadLine(), out second))
                    {
                        product = computeProd1(first, second);
                    }
                    else
                    {
                        Restart();
                    }
                }
                else
                {
                    Restart();
                }


                Console.WriteLine("Their product is:\t" + product);

                Console.ReadLine();// So it wont close.
            }
            catch(Exception ex)
            {
                throw ex;
            }
        }

        public static void Restart()
        {
            Console.WriteLine("Oops something went wrong thats not a number!");
            Console.WriteLine("Restarting in 3 seconds...");
            System.Threading.Thread.Sleep(3000);
            var fileName =   System.Reflection.Assembly.GetExecutingAssembly().Location;
            System.Diagnostics.Process.Start(fileName);
            Environment.Exit(0);
        }
    }
}

Answer 3

默认Console.Read()根据当前character encoding返回输入字符的值。例如，这里是windows-1250 的链接，它可能解释了为什么 0 得到 48 等。

如果您只想使用单个数字，您可以使用Convert.ToChar() 获取实际的键值。对于多位数值，您可以使用int.TryParse() 方法。以下是两者的示例：

int first, second;
//for single digit inputs
var singleDigitInput1 = Convert.ToChar(Console.Read).ToString();
var singleDigitInput2 = Convert.ToChar(Console.Read).ToString();

//for multi digit inputs
var multiDigitInput1 = Console.ReadLine();
var multiDigitInput2 = Console.ReadLine()

if (int.TryParse(singleDigitInput1, out first) && int.TryParse(singleDigitInput2, out second))
{
    int product = computeProd1(first, second);
    Console.WriteLine("Their product is:\t" + product);
}
else
{
    //input cannot be converted to int
}