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Function overloading

Function overloading happens when more than one function with the same name is declared with different parameters. This is a useful feature when an operation needs to be able to work with different types of input, yet produce similar result.

The compiler is able to resolve which function to call by matching the type of each argument expression to the function parameter, and eliminating the functions where there is no possible conversion available. The compiler will do this for each argument from first to last. When all arguments have been evaluated only one function should be the best match, otherwise the compiler will give an error about the impossibility to determine the correct function to call.

The type of conversion that needs to be performed on the argument to get to the type of the parameter determines how well a function matches. The following list gives the order of how one type of conversion compares to another.

  • no conversion needed
  • conversion to const
  • conversion of enum to integer of same size
  • conversion of enum to integer of different size
  • size of primitive type increases
  • size of primitive type decreases
  • signed integer to unsigned integer conversion
  • unsigned integer to signed integer conversion
  • integer type to float type
  • float type to integer type
  • reference cast
  • object to primitive conversion
  • conversion to object
  • variable argument type

Observe that it is not possible to create overloads where the only difference is the return type. This is because the return type is not part of the selection criteria that the compiler uses to determine which function to call, the return type is just the result of the called function.

  void Function(int a, float b, string c) {}
  void Function(string a, int b, float c) {}
  void Function(float a, string b, int c) {}

  void main()
  {
    Function(1, 2.5f, 'a');  // Will call the first overload
    Function('a', 1, 2.5f);  // Will call the second overload
    Function(2.5f, 'a', 1);  // Will call the third overload
  }