Inout parameter

 In Swift, functions can accept parameters as references using the "inout" keyword. When a parameter is passed as an inout parameter, any changes made to its value within the function are reflected in the original variable that was passed in. 

Here's an example of how to define a function that takes an inout parameter:

func increment(number: inout Int) {
    number += 1
}

In this example, the function "increment" takes an inout parameter called "number" of type "Int". This parameter is passed by reference using the "inout" keyword. 

To call the "increment" function and pass in a variable as an inout parameter, you must use the "&" operator to indicate that you are passing the variable by reference:

var myNumber = 5
increment(number: &myNumber)
print(myNumber) // Outputs 6

In this example, the "increment" function modifies the value of "myNumber" by adding 1 to it. Because "myNumber" was passed as an inout parameter, the original value of the variable is modified even though it was declared outside of the function.

It's important to note that not all parameters can be passed as inout parameters. For example, parameters that are constants or literals cannot be passed as inout parameters. Additionally, functions that take inout parameters cannot be used as closures or passed as arguments to functions that expect non-inout parameters.