In Swift, an inout parameter is a way to pass a parameter to a function or method by reference, allowing the function to modify the original value of the parameter. This is different from regular function parameters, which are passed by value, meaning a copy of the parameter's value is created within the function's scope, and any changes made to the parameter inside the function do not affect the original value outside of the function.

Here's how you declare and use an inout parameter in Swift:


func modifyValue(inout value: Int) {
value += 1
}

var number = 5
modifyValue(&number) // Pass the parameter with an ampersand "&" to indicate it's inout
print(number) // This will print 6 because the original value was modified inside the function


n the example above, the modifyValue function takes an inout parameter named value. When you call this function and pass number with an ampersand (&) before it, you're telling Swift to pass a reference to the original number variable, not a copy. Any changes made to value inside the function will directly affect the number variable outside of the function.

It's essential to use inout parameters carefully because they can make code less predictable and harder to reason about. They are typically used when you need a function to modify a variable's value and reflect that change outside of the function's scope.