当我尝试访问Codable
子类实例的属性并且满足以下两个条件之一时,我的应用程序崩溃:
- 子类是一个多级子类
Codable
- 有一个调用
JSONEncoder().encode
该函数的函数不必调用,它只需要出现在您实例化相关类的位置即可。
实体.swift:
class Entity: Codable {
}
ChildEntity.swift:
// If ChildEntity inherits from Entity, the app will crash
/* If ChildEntity simply implements Codable like so : class ChildEntity: Codable,
the app will not crash even with the 'causesCorruptionEvenIfNotCalled' function in ChildEntityController
*/
class ChildEntity: Entity {
var name: String = ""
}
ViewController.swift:(初始视图控制器)
import UIKit
class ViewController: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
let childEntity: ChildEntity = ChildEntity()
// Simply having the function 'causesCorruptionEvenIfNotCalled' in the code (not calling it) causes 'childEntity' properties to become inacessible
// Before accessing the property, 'childEntity.name' is a normal empty String
let name: String = childEntity.name // CRASH: name cannot be accessed because of a EXC_BAD_ACCESS error
// At this point 'childEntity.name' is corrupted with random data but 'name' has an empty String
// You can get get the property's value but inspecting either 'name' and 'childEntity.name' throws and EXC_BAD_ACCESS error
print("name: \(name)")
}
// Commenting or removing this function's body will prevent all the issues below
func causesCorruptionEvenIfNotCalled(object: Entity) {
let _: Data? = try? JSONEncoder().encode(object)
}
}
有几件事我很困惑:
只是有一个调用的函数
JSONEncoder().encode
会导致崩溃。即使没有在任何地方调用该函数。如果你
let _: Data? = try? JSONEncoder().encode(childEntity)
在初始化之后立即放ChildEntity
,应用程序不会崩溃,即使你让causesCorruptionEvenIfNotCalled
我刚才说的功能。如果
ChildEntity
直接继承自Codable
,则没有问题,应用程序不会崩溃。
如何在使用 JSON 编码器保持继承结构和功能的同时防止崩溃?
这是一个示例项目:https ://drive.google.com/open?id=1mrhOmm4kOAdMjLk5nlFLDeo6vTsBo1Uv