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Reflection.Emit. How to generate a simple if statement.

Posted By:      Posted Date: October 19, 2010    Points: 0   Category :.NET Framework

I am trying to generate a simple C# function that takes in an object as a parameter and outputs it through Console.WriteLine.

Something like this:

        public void DesiredMethod(object input) {
            if (input != null) {

I was able to generate similar function but without the IF statement that checks for null. Here is the code:


        private static object GenerateObj() {

            AssemblyBuilder assmBuilder = AppDomain.CurrentDomain.DefineDynamicAssembly(new System.Reflection.AssemblyName() { Name = "MyAssembly" },

            ModuleBuilder moduleBuilder = assmBuilder.DefineDynamicModule("MainModule", true);

            TypeBuilder typeBuilder = moduleBuilder.DefineType("MyType", T

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