How to assign value to this type in Swift?

Asked Jul 20 '15 at 06:03

Active Jul 20 '15 at 07:26

Viewed 70 times

I'm using Swift as my project language,and there is a library written in Objective-C.The issue is that the property apnsNickname's type is String!? in Swift.As the compiler's auto completion tells me.

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I want to know how to assign a value to this type.The following assignment has a compiler error.

EaseMob.sharedInstance().chatManager.apnsNickname = "Mike"

Cannot assign to the result of this expression

Can anyone help me out?

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The definition of the property is:

@protocol IChatManagerSettingOptions <IChatManagerBase>

@optional

@property (strong, nonatomic) NSString *apnsNickname;

...

@end

edited Jul 20 '15 at 06:22

asked Jul 20 '15 at 06:03

tounaobun

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"Mike" is actually of type `String`. – tounaobun Jul 20 '15 at 06:06
Still you need to typecasting! As swift is a "type safe" language – Sohil R. Memon Jul 20 '15 at 06:07
Also as you can see that there is "!" and "?" both, so it says that you specify that explicitly that it's a type of String or it will convert automatically – Sohil R. Memon Jul 20 '15 at 06:08
Can you show compile log? – Long Pham Jul 20 '15 at 06:14
@longpham The compiler log is `Cannot assign to the result of this expression`. – tounaobun Jul 20 '15 at 06:20
I think: `EaseMob.sharedInstance().chatManager.apnsNickname = "Mike"` is exact. – Long Pham Jul 20 '15 at 06:21
I don't think there's much sense or clearness in an optional property, and I can imagine this leading to problems. – Eiko Jul 20 '15 at 06:39
[This answer](http://stackoverflow.com/a/29925097/1187415) to the referenced thread shows a workaround which is also discussed in https://devforums.apple.com/message/994834. – Martin R Jul 20 '15 at 08:07

How to assign value to this type in Swift?

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