Question

我有一个带可选参数（位置）的函数。我测试它是否为零，但Xcode仍然显示错误：“可选类型Int的值？未解包”并建议我使用“！”或“？”。

var entries = [String]()

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

我是Swift的新手，不明白为什么这不好。在这个if子句中，编译器应该100％确定定义了位置，或者？

Answer 1

有几种方法可以正确编码：

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

或：

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
    }
}

Answer 2

即使您检查了nil，您仍然没有将Int?转换为Int以用作索引。为此，您需要打开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
}

Swift可选参数未解包

2 个答案: