Swift 4 - Cannot invoke 'encode' with an argument list of type '(Codable)'
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
add a comment |
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
You need to specify the exact type while usingCodable
. 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 get1. Expected an argument list of type '(T)'
which can be related to Dávid Pásztor's answer.
– Larme
Nov 13 '18 at 16:21
add a comment |
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
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
json swift codable encodable
asked Nov 13 '18 at 16:12
MarkCoolskiMarkCoolski
386
386
You need to specify the exact type while usingCodable
. 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 get1. Expected an argument list of type '(T)'
which can be related to Dávid Pásztor's answer.
– Larme
Nov 13 '18 at 16:21
add a comment |
You need to specify the exact type while usingCodable
. 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 get1. 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
add a comment |
1 Answer
1
active
oldest
votes
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)
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
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)
add a comment |
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)
add a comment |
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)
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)
answered Nov 13 '18 at 16:17
Dávid PásztorDávid Pásztor
22.1k82849
22.1k82849
add a comment |
add a comment |
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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