Swift Codable手动解码可选变量

我有以下代码：

import Foundation

let jsonData = """
[
    {"firstname": "Tom", "lastname": "Smith", "age": "28"},
    {"firstname": "Bob", "lastname": "Smith"}
]
""".data(using: .utf8)!

struct Person: Codable {
    let firstName, lastName: String
    let age: String?

    enum CodingKeys : String, CodingKey {
        case firstName = "firstname"
        case lastName = "lastname"
        case age
    }

    init(from decoder: Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        firstName = try values.decode(String.self, forKey: .firstName)
        lastName = try values.decode(String.self, forKey: .lastName)
        age = try values.decode(String.self, forKey: .age)
    }

}

let decoded = try JSONDecoder().decode([Person].self, from: jsonData)
print(decoded)

问题是它崩溃了age = try values.decode(String.self, forKey: .age)。当我取出该init功能时，它工作正常。错误是No value associated with key age (\"age\").。

关于如何使该可选项不存在的任何想法，在不存在时会崩溃？init由于其他原因，我也需要该函数，但仅举了一个简单的示例来说明发生了什么。