Swift 4 Programming Language, inout parameter is not working for FunctionType as Paramter

Question

This is the sample code from swift documentation. I am learning swift language, and I saw the functiontype as parameter, The sample code does not have inout keyword. But I am trying to use this with inout paramter, but the below sample is not working as expected.

https://docs.swift.org/swift-book/LanguageGuide/Functions.html (Function Types as Return Types)

//Function Types as Return Types
func stepForward(_ input: inout Int) -> Int {
    return input + 1
}
func stepBackward(_ input: inout Int) -> Int {
    return input - 1
}
func chooseStepFunction(backward: Bool) -> (inout Int) -> Int {
    let a = backward ? stepBackward : stepForward
    return a
}
var currentValue = 3
let moveNearerToZero = chooseStepFunction(backward: currentValue > 0)
print(moveNearerToZero(&currentValue))
print(currentValue)

Actual output 2 3

Expected output 2 2

Because CurrentValue is inout paramter. Passing the currentValue as 3 initially prints value 2 using stepBackward() method

and I want to maintain the value after the decrement.

But the currentValue is not maintained here.

Answer 1

That's because you are not actually assigning value to parameter after applying arithmetics you are just returning new value without assigning it. Try the following code

//Function Types as Return Types
func stepForward(_ input: inout Int) -> Int {
    input += 1
    return  input
}
func stepBackward(_ input: inout Int) -> Int {
    input -= 1
    return  input 
}
func chooseStepFunction(backward: Bool) -> (inout Int) -> Int {
    let a = backward ? stepBackward : stepForward
    return a
}
var currentValue = 3
let moveNearerToZero = chooseStepFunction(backward: currentValue > 0)
print(moveNearerToZero(&currentValue))
print(currentValue)