Swift 4 - Cannot invoke 'encode' with an argument list of type '(Codable)'










1















I have built a set of API functions which encode an object (using a Struct which conforms to Codable), then Posts the resulting JSON Data object to a server, then decodes the JSON response. All works fine - especially happy with the new method for JSON parsing in Swift 4.2. However, now I want to refactor the code so that I can reuse the code for various method calls - when I do I get a really annoying error.



func encodeRequestJSON(apiRequestObject: Codable) -> Data 
do
let encoder = JSONEncoder()
let jsonData = try encoder.encode(apiRequestObject)
let jsonString = String(data: jsonData, encoding: .utf8)
print(jsonString)
catch
print("Unexpected error")

return jsonData!



This is the error message:



Cannot invoke 'encode' with an argument list of type '(Codable)'


I have tried changing the type from Codable, to Encodable but get the same error, except with type (Encodable) in the message. Any advice? My fall-back is to encode the data in the current ViewController, then call the HTTPPost function and then decode back in the VC. But that's really clunky.










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  • You need to specify the exact type while using Codable. You can't generalize it with protocol as a type.

    – PGDev
    Nov 13 '18 at 16:17











  • Note that if you click on the "more info" of the error, you should get 1. Expected an argument list of type '(T)' which can be related to Dávid Pásztor's answer.

    – Larme
    Nov 13 '18 at 16:21















1















I have built a set of API functions which encode an object (using a Struct which conforms to Codable), then Posts the resulting JSON Data object to a server, then decodes the JSON response. All works fine - especially happy with the new method for JSON parsing in Swift 4.2. However, now I want to refactor the code so that I can reuse the code for various method calls - when I do I get a really annoying error.



func encodeRequestJSON(apiRequestObject: Codable) -> Data 
do
let encoder = JSONEncoder()
let jsonData = try encoder.encode(apiRequestObject)
let jsonString = String(data: jsonData, encoding: .utf8)
print(jsonString)
catch
print("Unexpected error")

return jsonData!



This is the error message:



Cannot invoke 'encode' with an argument list of type '(Codable)'


I have tried changing the type from Codable, to Encodable but get the same error, except with type (Encodable) in the message. Any advice? My fall-back is to encode the data in the current ViewController, then call the HTTPPost function and then decode back in the VC. But that's really clunky.










share|improve this question






















  • You need to specify the exact type while using Codable. You can't generalize it with protocol as a type.

    – PGDev
    Nov 13 '18 at 16:17











  • Note that if you click on the "more info" of the error, you should get 1. Expected an argument list of type '(T)' which can be related to Dávid Pásztor's answer.

    – Larme
    Nov 13 '18 at 16:21













1












1








1








I have built a set of API functions which encode an object (using a Struct which conforms to Codable), then Posts the resulting JSON Data object to a server, then decodes the JSON response. All works fine - especially happy with the new method for JSON parsing in Swift 4.2. However, now I want to refactor the code so that I can reuse the code for various method calls - when I do I get a really annoying error.



func encodeRequestJSON(apiRequestObject: Codable) -> Data 
do
let encoder = JSONEncoder()
let jsonData = try encoder.encode(apiRequestObject)
let jsonString = String(data: jsonData, encoding: .utf8)
print(jsonString)
catch
print("Unexpected error")

return jsonData!



This is the error message:



Cannot invoke 'encode' with an argument list of type '(Codable)'


I have tried changing the type from Codable, to Encodable but get the same error, except with type (Encodable) in the message. Any advice? My fall-back is to encode the data in the current ViewController, then call the HTTPPost function and then decode back in the VC. But that's really clunky.










share|improve this question














I have built a set of API functions which encode an object (using a Struct which conforms to Codable), then Posts the resulting JSON Data object to a server, then decodes the JSON response. All works fine - especially happy with the new method for JSON parsing in Swift 4.2. However, now I want to refactor the code so that I can reuse the code for various method calls - when I do I get a really annoying error.



func encodeRequestJSON(apiRequestObject: Codable) -> Data 
do
let encoder = JSONEncoder()
let jsonData = try encoder.encode(apiRequestObject)
let jsonString = String(data: jsonData, encoding: .utf8)
print(jsonString)
catch
print("Unexpected error")

return jsonData!



This is the error message:



Cannot invoke 'encode' with an argument list of type '(Codable)'


I have tried changing the type from Codable, to Encodable but get the same error, except with type (Encodable) in the message. Any advice? My fall-back is to encode the data in the current ViewController, then call the HTTPPost function and then decode back in the VC. But that's really clunky.







json swift codable encodable






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asked Nov 13 '18 at 16:12









MarkCoolskiMarkCoolski

386




386












  • You need to specify the exact type while using Codable. You can't generalize it with protocol as a type.

    – PGDev
    Nov 13 '18 at 16:17











  • Note that if you click on the "more info" of the error, you should get 1. Expected an argument list of type '(T)' which can be related to Dávid Pásztor's answer.

    – Larme
    Nov 13 '18 at 16:21

















  • You need to specify the exact type while using Codable. You can't generalize it with protocol as a type.

    – PGDev
    Nov 13 '18 at 16:17











  • Note that if you click on the "more info" of the error, you should get 1. Expected an argument list of type '(T)' which can be related to Dávid Pásztor's answer.

    – Larme
    Nov 13 '18 at 16:21
















You need to specify the exact type while using Codable. You can't generalize it with protocol as a type.

– PGDev
Nov 13 '18 at 16:17





You need to specify the exact type while using Codable. You can't generalize it with protocol as a type.

– PGDev
Nov 13 '18 at 16:17













Note that if you click on the "more info" of the error, you should get 1. Expected an argument list of type '(T)' which can be related to Dávid Pásztor's answer.

– Larme
Nov 13 '18 at 16:21





Note that if you click on the "more info" of the error, you should get 1. Expected an argument list of type '(T)' which can be related to Dávid Pásztor's answer.

– Larme
Nov 13 '18 at 16:21












1 Answer
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You need a concrete type to be passed into JSONEncoder.encode, so you need to make your function generic with a type constraint on Encodable (Codable is not needed, its too restrictive).



func encodeRequestJSON<T:Encodable>(apiRequestObject: T) throws -> Data 
return try JSONEncoder().encode(apiRequestObject)






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    You need a concrete type to be passed into JSONEncoder.encode, so you need to make your function generic with a type constraint on Encodable (Codable is not needed, its too restrictive).



    func encodeRequestJSON<T:Encodable>(apiRequestObject: T) throws -> Data 
    return try JSONEncoder().encode(apiRequestObject)






    share|improve this answer



























      3














      You need a concrete type to be passed into JSONEncoder.encode, so you need to make your function generic with a type constraint on Encodable (Codable is not needed, its too restrictive).



      func encodeRequestJSON<T:Encodable>(apiRequestObject: T) throws -> Data 
      return try JSONEncoder().encode(apiRequestObject)






      share|improve this answer

























        3












        3








        3







        You need a concrete type to be passed into JSONEncoder.encode, so you need to make your function generic with a type constraint on Encodable (Codable is not needed, its too restrictive).



        func encodeRequestJSON<T:Encodable>(apiRequestObject: T) throws -> Data 
        return try JSONEncoder().encode(apiRequestObject)






        share|improve this answer













        You need a concrete type to be passed into JSONEncoder.encode, so you need to make your function generic with a type constraint on Encodable (Codable is not needed, its too restrictive).



        func encodeRequestJSON<T:Encodable>(apiRequestObject: T) throws -> Data 
        return try JSONEncoder().encode(apiRequestObject)







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 13 '18 at 16:17









        Dávid PásztorDávid Pásztor

        22.1k82849




        22.1k82849





























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