Question

class foo
{
  public void bar(int i) { ... };
  public void bar(long i) { ... };
}


foo.bar(10);

I would expect this code to give me some error, or at least an warning, but not so…

What version of bar() is called, and why?

Answer 1

The int version of bar is being called, because 10 is an int literal and the compiler will look for the method which closest matches the input variable(s). To call the long version, you’ll need to specify a long literal like so: foo.bar(10L);

Here is a post by Eric Lippert on much more complicated versions of method overloading. I’d try and explain it, but he does a much better job and I ever could: http://blogs.msdn.com/b/ericlippert/archive/2006/04/05/odious-ambiguous-overloads-part-one.aspx

from the C# 4.0 Specification:

Method overloading permits multiple
methods in the same class to have the
same name as long as they have unique
signatures. When compiling an
invocation of an overloaded method,
the compiler uses overload resolution
to determine the specific method to
invoke. Overload resolution finds the
one method that best matches the
arguments or reports an error if no
single best match can be found. The
following example shows overload
resolution in effect. The comment for
each invocation in the Main method
shows which method is actually
invoked.

 class Test {   
      static void F() {
        Console.WriteLine("F()");   
      }     
      static void F(object x) {
        Console.WriteLine("F(object)");     
      }
      static void F(int x) {
        Console.WriteLine("F(int)");    
      }
      static void F(double x) {
        Console.WriteLine("F(double)");     
      }
      static void F<T>(T x) {
        Console.WriteLine("F<T>(T)");   
      }
      static void F(double x, double y) {
        Console.WriteLine("F(double,double)");  
      }     

      static void Main() {
        F();                // Invokes F()
        F(1);           // Invokes F(int)
        F(1.0);         // Invokes F(double)
        F("abc");       // Invokes F(object)
        F((double)1);       // Invokes F(double)
        F((object)1);       // Invokes F(object)
        F<int>(1);      // Invokes F<T>(T)
        F(1, 1);        // Invokes F(double, double)
      } 
}

As shown by the example, a particular
method can always be selected by
explicitly casting the arguments to
the exact parameter types and/or
explicitly supplying type arguments.