Question

这是我的代码：我已经有了一个名为“Rooms”的对象

class BuildRooms  {
func build(room: [Rooms]){
    room[0].setExits([-1,-1,-1,-1])
}

我试图通过输入以下来使用它：

var room: [Rooms] = []
BuildRooms.build(room)

BuildRooms.build（房间）发生错误它说：无法将'Rooms'类型的值转换为'BuildRooms'的预期参数类型

感谢

Answer 1

您尝试将函数构建称为类/或静态/函数，但它被声明为实例函数。我同意，您收到的错误消息可能令人困惑，但这是正确的。

class Rooms {}
class BuildRooms  {
    class func build(room: [Rooms]){
        print("room: \(room)")
    }
}

var room: [Rooms] = []
BuildRooms.build(room) // room: []

你做了什么......可以这样做

class Rooms {}
class BuildRooms  {
    func build(room: [Rooms]){
        print("room: \(room)")
    }
}

var room: [Rooms] = []
let br = BuildRooms()
let bf = BuildRooms.build(br) // here the parameter expected is concrete instance of BuildRooms class and returning type is ...
print(bf.dynamicType) // Array<Rooms> -> ()
bf(room) // room: []

Swift 2：使用对象数组作为参数调用函数

1 个答案: