Swift optional parameter not unwrapped

问题

I have a function with a optional parameter(position). I test for it to be nil but still Xcode shows me an error: "Value of optional type Int? not unwrapped" and suggests me to use "!" or "?".

var entries = [String]()

func addEntry(text: String, position: Int?) {
    if(position == nil) {
        entries.append(text)
    } else {
        entries[position] = text
    }
}

Im new to Swift and don't understand why this isn't ok. Within this if-clause the compiler should be 100% sure that position is defined, or?

回答1:

There are a few ways to code this properly:

func addEntry(text: String, position: Int?) {
    // Safely unwrap the value
    if let position = position {
        entries[position] = text
    } else {
        entries.append(text)
    }
}

or:

func addEntry(text: String, position: Int?) {
    if position == nil {
        entries.append(text)
    } else {
        // Force unwrap since you know it isn't nil
        entries[position!] = text
    }
}

回答2:

Even though you checked for nil you still haven’t converted the Int? into an Int to use as an index. To do that you need to unwrap the Optional:

var entries = [String]()

func addEntry(text: String, position: Int?) {
  // guard against a nil position by attempting to
  // unwrap it with optional binding
  guard
    // test for nil and unwrap using optional binding
    let unwrappedPosition = position,
    // make sure the position isn't 
    // past the end of the array
    unwrappedPosition < entries.count else {
      // can't unwrap it or 
      // it's out of bounds,
      // append it
      entries.append(text)
      return
  }

  // successfully unwrapped, set item at index
  entries[unwrappedPosition] = text
}

来源：https://stackoverflow.com/questions/40290202/swift-optional-parameter-not-unwrapped