Swift 4：使用Codable`通用参数'T'无法推断

Question

I am getting the following error:

Generic parameter 'T' could not be inferred

on line: let data = try encoder.encode(obj)

Here is the code

import Foundation

struct User: Codable {
    var firstName: String
    var lastName: String
}

let u1 = User(firstName: "Ann", lastName: "A")
let u2 = User(firstName: "Ben", lastName: "B")
let u3 = User(firstName: "Charlie", lastName: "C")
let u4 = User(firstName: "David", lastName: "D")

let a = [u1, u2, u3, u4]

var ret = [[String: Any]]()
for i in 0..<a.count {
        let param = [
            "a" : a[i],
            "b" : 45
            ] as [String : Any]
    ret.append(param)
}


let obj = ["obj": ret]

let encoder = JSONEncoder()
encoder.outputFormatting = .prettyPrinted
let data = try encoder.encode(obj) // This line produces an error
print(String(data: data, encoding: .utf8)!)

What am I doing wrong?

Answer 1

That message is misleading, the real error is that obj is of type [String: Any] , which does not conform to Codable because Any does not.

When you think about it, Any can never conform to Codable . What will Swift use to store the JSON entity when it can be an integer, a string, or an object? You should define a proper struct to hold your data.