C# 如何修复最佳重载方法匹配有一些无效参数？
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原文地址: http://stackoverflow.com/questions/15703737/
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How to fix the best overloaded method match has some invalid arguments?
提问by user2224223
I am getting the error in the title, what is wrong with the code? I think its a syntax error, but I'm not sure as I don't have much information on what the error actually means.
我收到标题中的错误，代码有什么问题？我认为这是一个语法错误，但我不确定，因为我没有太多关于错误实际含义的信息。
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace ConsoleApplication2
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Please Input Number of Rows you want to make in your pyrimid: ");
int num = int.Parse(Console.Read()); // error here
Console.WriteLine(num);// Just to check if it is getting the right number
Console.Read();//This is Here just so the console window doesn't close when the program runs
}
}
}
Edit:
编辑：
Just to Clarify, I want the code just to get a number from the user and then print the number the user inputted.
只是为了澄清，我希望代码只是从用户那里获取一个数字，然后打印用户输入的数字。
采纳答案by Ilya Ivanov
int.Parseaccepts a string as parameter. Use Console.ReadLine()to get a string from user and then pass it into int.Parse
int.Parse接受一个字符串作为参数。用于Console.ReadLine()从用户获取字符串，然后将其传递给int.Parse
int num = int.Parse(Console.ReadLine());
Note that this will throw FormatExceptionif the user will enter something not recognisable as int. If you are not sure, that the user will input a good number (I always don't), use TryParse. Example is given below
请注意，FormatException如果用户输入无法识别的内容，这将抛出int。如果您不确定用户是否会输入一个好的数字（我总是不这样做），请使用TryParse. 示例如下
int value;
if (int.TryParse(Console.ReadLine(), out value))
Console.WriteLine("parsed number as: {0}", value);
else
Console.WriteLine("incorrect number format");
回答by VladL
The problem is that Console.Read() returns an int, but int.Parse is expecting a string. Simply change it to
问题是 Console.Read() 返回一个 int，但 int.Parse 需要一个字符串。只需将其更改为
int num =Console.Read();
回答by Haedrian
The reason you're getting that, is because Console.Read returns an int
你得到它的原因是因为 Console.Read 返回一个 int
http://msdn.microsoft.com/en-us/library/system.console.read.aspx
http://msdn.microsoft.com/en-us/library/system.console.read.aspx
And it can't parse an int, it can only parse strings.
它不能解析 int，它只能解析字符串。
You probabily want Console.ReadLine - which returns a string.
您可能需要 Console.ReadLine - 它返回一个字符串。
回答by Daniel Imms
Console.Read()and ASCII
Console.Read()和 ASCII
This is occurring because Console.Read()actually returns and int, not a string. It returns the ASCII code of the key that was pressed, you will need to convert it to a char then to a string and then parse it.
发生这种情况是因为Console.Read()实际上返回的是 and int，而不是 a string。它返回按下的键的 ASCII 代码，您需要将其转换为字符，然后转换为字符串，然后解析它。
var val = int.Parse(((char)Console.Read()).ToString());
Note that Console.Read()will not return the integer in the format you think, values 0to 9actually come out at 60to 70as they're the key codesnot the characters you pressed.
请注意，Console.Read()不会在你认为的格式返回整，价值观0，以9实际的出来60，以70作为他们的键码不是你按下的字符。
Console.ReadLine()
Console.ReadLine()
An alternative and probably better solution would be to use Console.ReadLine()which returns a string
另一种可能更好的解决方案是使用Console.ReadLine()which 返回一个string
var val = int.Parse(Console.ReadLine());
Warning
警告
You should always be careful when using int.Parse()as it will throw an exception if the string provided is not numeric. A better option is to use int.TryParse()which you give an outargument and it returns whether parsing was successful.
使用时应始终小心，int.Parse()因为如果提供的字符串不是数字，它将引发异常。更好的选择是使用int.TryParse()您给出的out参数，它返回解析是否成功。
string text = Console.ReadLine();
int val;
if (int.TryParse(text, out val))
{
// It is a number
}
{
// It is not a number
}
