如何在词典中迅速分配到AnyObject

问题描述

我一直在玩弄斯威夫特。我曾与多个类型的错误，尤其是斯威夫特和我的老Objective-C类的工作。这种方法的问题是：我期待在Objective-C制成NSDictionarys数组

I have been playing around with Swift. I have had multiple errors with types, especially working with Swift and my old Objective-C classes. The problem with this method is: I am expecting an array made of NSDictionarys in Objective-C.

var curArr:[Dictionary<String, AnyObject>] = self.getItemsForCurrentStack()
var arrToReturn:[(Dictionary<String, AnyObject?>)] = []
for obj in curArr{
    arrToReturn.append(["image": UIImage(named: obj["imageName"] as! String), "color": UIColor(red:obj["colorDic"]!["red"] as! CGFloat, green:obj["colorDic"]!["green"] as! CGFloat, blue:obj["colorDic"]!["blue"] as! CGFloat, alpha:1.0), "percentage": obj["percantage"]])
}
return arrToReturn

这将返回斯威夫特字典（这是NSDictionaries）。但最后一行抛出我一个错误：

This returns Dictionaries (which are NSDictionaries) in Swift. But the last line throws me an error:

字典不是完全相同'AnyObject'

我已经使用的尝试！ [AnyObject]

但是，这将引发另一个错误：

But that throws another error:

AnyObject不是字典

我没有得到第二误差，因为这并不必须是一个亚型，但周围的其他方法。任何想法如何解决此问题？我没有找到答案了几个小时googleing和研究的。

I don't get the second error, since this doesn't have to be a subtype, but the other way around. Any ideas on how to solve this? I didn't find an answer for several hours of googleing and researching.

推荐答案

词典是斯威夫特一个结构，而 AnyObject 是

Dictionary is a struct in Swift, whereas AnyObject is

///协议，这是所有类都隐式遵循。

根据你想要做什么，你可能需要使用任何在code或投你的字典NSDictionary中使用作为的NSDictionary 。

Depending what you're trying to do, you may want to use Any in your code, or cast your dictionary to NSDictionary using as NSDictionary.

你的澄清后编辑：

如果你分割了从字典本身追加通话，你看到一个更好的错误消息：

If you split up the append call from the dictionary itself, you see a better error message:

所以，问题是，你的字典包含一些可选值，但可选的是一个结构，而不是转换为的OBJ-C。您可以通过转换成 UIImage的修复！和 AnyObject！（ImplicitlyUnwrappedOptional），或使用如！。

So, the issue is that your dictionary contains some Optional values, but Optional is a struct and not convertible to Obj-C. You can fix by casting to UIImage! and AnyObject! (ImplicitlyUnwrappedOptional), or by using as!.

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