Swift with() and withLet() functions

WithFunctions.swift

/**
 Groups operations on an item, using shorthand to reduce repetition.

 Using this function, repetitive code like
 ```
 let myWellDescribedLabel = UILabel()
 myWellDescribedLabel.attributedText = attributedStringTitle
 myWellDescribedLabel.isAccessibilityElement = true
 myWellDescribedLabel.numberOfLines = 1
 ```
 can be replaced with
 ```
 let myWellDescribedLabel = with(UILabel()) {
   $0.attributedText = attributedStringTitle
   $0.isAccessibilityElement = true
   $0.numberOfLines = 1
 }
 ```

 - Parameters:
   - item: item to operate on, either a reference or value type
   - closure: closure operating on that item

 - Returns: item after applying closure
 */
@discardableResult
public func with<T>(_ item: T, _ closure: (inout T) -> Void) -> T {
  var mutableItem = item
  closure(&mutableItem)
  return mutableItem
}

/**
 Groups operations on an *_optional_* item, using shorthand to reduce repetition.

 This function is the optional-handling version of `with()`. Much like an if-let, the closure
 is run only for non-nil items, and in the scope of the closure, the item is not optional
 (e.g. referenced as `$0` and not `$0?`).

 - Parameters:
   - item: optional item to operate on, either a reference or value type
   - closure: closure operating on that item

 - Returns: item after applying closure
 */
@discardableResult
public func withLet<T>(_ item: Optional<T>, _ closure: (inout T) -> Void) -> Optional<T> {
  guard let item = item else { return nil }
  return with(item, closure)
}