I am confused on the "out" keyword in the TryParse method.

out is a keyword that modifies how the argument is passed into the method, and it is considered part of the method's signature. Meaning that the keyword has to be included both when the method is defined and when it is called.

It causes the argument to be passed by reference, and not by value as is usually the case.

Take this code for example:

using System;

class MainClass {
  public static void Main()
   {
      int var1 = 0;
      SetVariableTo5(var1);
      Console.WriteLine(var1);
   }

   public static void SetVariableTo5(int a)
   {
      a = 5;
   }
}

What would you expect the above code to print? Well the answer is 0, because while you pass var1 into SetVariableTo5 that only passed the value of var1, not the actual variable so changing the value of a won't do anything to var1.

With this code however:

using System;

class MainClass {
  public static void Main()
   {
      int var1 = 0;
      SetVariableTo5(out var1);
      Console.WriteLine(var1);
   }

   public static void SetVariableTo5(out int a)
   {
      a = 5;
   }
}

Things are different. With this code 5 will in fact be printed. And this is due to the fact that when out is used you pass in the actual variable to the function, not just its value. Which means any changes done to it within the method will affect the variable that was passed in as an argument.

It's worth mentioning that there is also a keyword called ref that works pretty much identically to out, with the main difference being that ref requires that the variable has to have a value assigned to it. out accepts empty variables.

So code like this:

using System;

class MainClass {
  public static void Main()
   {
      int var1;
      SetVariableTo5(out var1);
      Console.WriteLine(var1);
   }

   public static void SetVariableTo5(out int a)
   {
      a = 5;
   }
}

Is valid with the out keyword but not with the ref keyword. While the code shown in the other example I provided would work with either out or ref.