Question

I want to know that how the methods are invoked using Reflection in c# .

Here is the code i have picked from book

using System;
using System.Reflection;
class MyClass {
int x;
int y;
public MyClass(int i, int j) {
x = i;
y = j;
}
public int Sum() {
return x+y;
}
public bool IsBetween(int i) {
if((x < i) && (i < y)) return true;
else return false;
}
public void Set(int a, int b) {
Console.Write("Inside Set(int, int). ");
x = a;
      y = b;
Show();
}
// Overload set.
public void Set(double a, double b) {
Console.Write("Inside Set(double, double). ");
x = (int) a;
y = (int) b;
Show();
}
public void Show() {
Console.WriteLine("Values are x: {0}, y: {1}", x, y);
}
public void Show1() {
Console.WriteLine("Hello");
} 
}
class InvokeMethDemo {
static void Main() {
Type t = typeof(MyClass);
MyClass reflectOb = new MyClass(10, 20);
int val;
Console.WriteLine("Invoking methods in " + t.Name);
Console.WriteLine();
MethodInfo[] mi = t.GetMethods();
// Invoke each method.
foreach(MethodInfo m in mi) {
// Get the parameters.
ParameterInfo[] pi = m.GetParameters();
if(m.Name.Equals("Set", StringComparison.Ordinal) &&
pi[0].ParameterType == typeof(int)) {
object[] args = new object[2];
args[0] = 9;
args[1] = 18;
m.Invoke(reflectOb, args);
}
else if(m.Name.Equals("Set", StringComparison.Ordinal) &&
pi[0].ParameterType == typeof(double)) {
object[] args = new object[2];
args[0] = 1.12;
args[1] = 23.4;
m.Invoke(reflectOb, args);
}
else if(m.Name.Equals("Sum", StringComparison.Ordinal)) {
val = (int) m.Invoke(reflectOb, null);
Console.WriteLine("sum is " + val);
}
else if(m.Name.Equals("IsBetween", StringComparison.Ordinal)) {
object[] args = new object[1];
args[0] = 14;
if((bool) m.Invoke(reflectOb, args))
Console.WriteLine("14 is between x and y");
}
else if(m.Name.Equals("Show", StringComparison.Ordinal)) {
m.Invoke(reflectOb, null);

}
else if(m.Name.Equals("Show1", StringComparison.Ordinal)) {
m.Invoke(reflectOb, null);
}         
}
}
}

1. Is the method invoked based on the number and type of parameters passed.If yes then Why

the Compiler does not show error as the **same statement is used to call method show and

show1**?

OR

1. The method is invoked based on some reference value of m(MethodInfo m) and the
   number of parameters.

Answer 1

You can invoke the Methods using MethodBase Class.

Invokes the method or constructor represented by the current instance, using the specified parameters.
Refer this link Method Invoking in Reflection