如何从我的方法中存储一些特定数据

Question

My program is having some methods and some of them are call some other methods my problem is that i want to use some data that a method is generating in a previous method and i dont know how im supposed to do that.

namespace MyProg
{
   public partial class MyProg: Form
   {
        public static void method1(string text)
        {
           //procedures
           method2("some text");
           // Here i want to use the value from string letter from method2.
        }

        public static void method2(string text)
        {
           //procedures
           string letter = "A";  //Its not A its based on another method.
        }      
   }
}

Answer 1

Just use return values:

public partial class MyProg: Form
{
    public static void method1(string text)
    {
       string letter = method2("some text");
       // Here i want to use the value from string letter from method2.
    }

    public static string method2(string text)
    {
       string letter = "A";  //Its not A its based on another method.
       return letter;
    }      
}

Methods

Methods can return a value to the caller. If the return type, the type listed before the method name, is not void, then the method can return the value using the return keyword. A statement with the keyword return followed by a value that matches the return type will return that value to the method caller...

---

Since you've mentioned that you cannot use return values, another option is to use an out parameter .

public static void method1(string text)
{
   string letter;
   method2("some text", out letter);
   // now letter is "A"
}

public static void method2(string text, out string letter)
{
   // ...
   letter = "A";
}  

Answer 2

You can either store the value in a member variable of the class (which in this case must be static since the method referring to it is static), or you can return the value from method2, and call method2 from inside method1 where you want to use it.

I'll leave it up to you to figure out how to code it.