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Home/ Questions/Q 9219661
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T03:09:59+00:00 2026-06-18T03:09:59+00:00

Say I am working with the following code: Type type = info.ParameterType; object activatedTypeToReference

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Say I am working with the following code:

Type type = info.ParameterType;
object activatedTypeToReference = Activator.CreateInstance(type.GetElementType());

How do I create a reference parameter object to the above activatedTypeToReference object in C#?

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  1. Editorial Team
    Editorial Team
    2026-06-18T03:10:00+00:00Added an answer on June 18, 2026 at 3:10 am

    When you invoke the method, you pass in an array of arguments. For an out parameter, you don’t need to specify anything for the array element – the value can just be null. When the method returns, the array will contain the value set by the method. Here’s an example:

    using System;
    
    public class Test
    {
        static void Main()
        {        
            var method = typeof(Test).GetMethod("DummyMethod");
            object[] args = new object[1];
            method.Invoke(null, args);
            Console.WriteLine(args[0]); // Prints 10
        }
    
        public static void DummyMethod(out int x)
        {
            x = 10;
        }
    }
    
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