Value of type 'T' has no dynamic member 'object' using key path from root type 'T'

Question

I had a class like this:

class ObjectA:ObservableObject { 
  @Published var type:ObjectSheetType = .none
}

and I was using a class like this to process instances of this kind of object:

class ProcessObject {

  @ObservedObject private var object:ObjectA

  init(object:ObjectA) {
    self.object = object
  }

  process() {
    if object.type == .big {
     // bla bla
    }
    ...
  }

Then I have decided to add another class:

class ObjectB:ObservableObject { 
  @Published var type:ObjectSheetType = .none
}

and a protocol

protocol ObjectProtocol {
  var type:ObjectSheetType { get set }
}

This protocol was added to both classes: ObjectA and ObjectB.

Then the class ProcessObject was modified to accept objects of both kinds:

class processObject<T:ObjectProtocol, ObservableObject > {

  @ObservedObject private var object:T

  init(object:T) {
    self.object = object
  }

Error: Referencing subscript 'subscript(dynamicMember:)' requires wrapper 'ObservedObject.Wrapper' Insert '$' Value of type 'T' has no dynamic member 'object' using key path from root type 'T'

error pointing to

if object.type == .big {

but also Xcode wants me to use $ in front of every published properties of object?

How do I solve this?

Does this answer your question https://stackoverflow.com/a/59504489/12299030? BTW @ObservedObject has no sense inside class, it is valid only inside SwiftUI view. — Asperi, Jul 09 '21 at 12:14
May be helpful for somebody, as of xcode14.1, this error is also returned if there is a typo in the variable name. In the example above `if object.types == .big {` returns this extremely confusing error... — jalone, Apr 19 '23 at 11:51

score 3 · Accepted Answer · answered Jul 09 '21 at 11:47

3

All you need to do is to fix the protocol conformance of T because now ObservableObject is seen as another generic type and not the protocol.

Use & to say that T should conform to both protocols

class ProcessObject<T: ObjectProtocol & ObservableObject> {

}

Another way is to say that ObjectProtocol inherits from ObservableObject and then only use ObjectProtocol in all the conform declarations

protocol ObjectProtocol: ObservableObject {
    var type:ObjectSheetType { get set }
}

but that changes what ObjectProtocol is and you may not want that tight coupling between the two protocols

answered Jul 09 '21 at 11:47

Joakim Danielson

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Thanks. Can you explain your last phrase: `but that changes what ObjectProtocol is and you may not want that tight coupling between the two protocols` ? – Duck Jul 09 '21 at 13:07
I always thought the comma on the <> was saying both. – Duck Jul 09 '21 at 13:10

Value of type 'T' has no dynamic member 'object' using key path from root type 'T'

1 Answers1