Question

我经历了一些类似的问题，但仍然无法弄清楚为什么会发生这种情况。

var liquors = [Liquor]()

func loadSampleLiquors(){
    let photo1 = UIImage(named: "Chateau Lafite Rothschild 1993")
    let liquor1 = Liquor(name: "Chateau Lafite Rothschild", year: "1993", photo: photo1, rating: 7)
    liquors += [liquor1] // Here is the error happen     
}

错误信息：无法将'[Liquor]'类型的值转换为预期的参数类型'inout _'

这可能是因为“年份”可能是零，但我通过我的代码它应该运行良好，我尝试使用“如果让liquors = xxx”修复它，但是那时会有一个EXC_BAD_INSTRUCTION解码功能，所以我在这里发布我的所有代码：

这是我的酒类：

var name: String
var year: String
var photo: UIImage?
var rating: Int

struct PropertyKey {
    static let nameKey = "name"
    static let yearKey = "year"
    static let photoKey = "photo"
    static let ratingKey = "rating"
}

我使用NSCoding存储数据：

func encode(with aCoder: NSCoder) {
    aCoder.encode(name, forKey: PropertyKey.nameKey)
    aCoder.encode(year, forKey: PropertyKey.yearKey)
    aCoder.encode(photo, forKey: PropertyKey.photoKey)
    aCoder.encode(rating, forKey: PropertyKey.ratingKey)
}

required convenience init?(coder aDecoder: NSCoder) {
    let name = aDecoder.decodeObject(forKey: PropertyKey.nameKey) as! String
    let year = aDecoder.decodeObject(forKey: PropertyKey.yearKey) as! String
    let photo = aDecoder.decodeObject(forKey: PropertyKey.photoKey) as? UIImage
    let rating = aDecoder.decodeInteger(forKey: PropertyKey.ratingKey)
    self.init(name: name, year:year, photo: photo, rating: rating)
}

Answer 1

您错过了解包操作员。试试这个：

let liquor1 = Liquor(name: "Chateau Lafite Rothschild", year: "1993", photo: photo1, rating: 7)!

注意＆＃34;！＆＃34;最后

你需要解包的原因是因为Liquor init可以返回nil，所以它返回一个可选项。多数民众赞成在？在init结束时

required convenience init?

swift无法将'[Liquor]'类型的值转换为预期的参数类型'inout _'

1 个答案: