C# 4 and .NET 4

Generic Methods ❘ 127
match is taken. If an int is passed, then the method with the int parameter is selected. With any other
parameter type, the compiler chooses the generic version of the method:
public class MethodOverloads
{
public void Foo(T obj)
{
Console.WriteLine("Foo(T obj), obj type: {0}", obj.GetType().Name);
}
public void Foo(int x)
{
Console.WriteLine("Foo(int x)");
}
}
public void Bar(T obj)
{
Foo(obj);
}
code snippet Specialization/Program.cs
The Foo() method can now be invoked with any parameter type. The sample code passes an int and a
string to the method:
static void Main()
{
var test = new MethodOverloads();
test.Foo(33);
test.Foo("abc");
}
code snippet GenericMethods/Program.cs
Running the program, you can see by the output that the method with the best match is taken:
Foo(int x)
Foo(T obj), obj type: String
You need to be aware that the method invoked is defined during compile time and not runtime. This can be
easily demonstrated by adding a generic Bar() method that invokes the Foo() method, passing the generic
parameter value along:
public class MethodOverloads
{
// ...
public void Bar(T obj)
{
Foo(obj);
}
The Main() method is now changed to invoke the Bar() method passing an int value:
static void Main()
{
var test = new MethodOverloads();
test.Bar(44);
From the output on the console you can see that the generic Foo() method was selected by the Bar()
method and not the overload with the int parameter. The reason is that the compiler selects the method
that is invoked by the Bar() method during compile time. Because the Bar() method defines a generic
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