我正在使用包含应用的iOS键盘扩展程序。我的问题是我无法“启动”键盘不会出现。框架是空的，我无法进入我的代码。这是为什么？ 我的设置是一个非常标准的键盘扩展，没有什么花哨的。我有一个托管应用程序，显示一些设置。我将此设置存储在NSUserDefaults套件中，并将我的应用程序组作为标识符。我是否需要完全访问才能访问此套件？
贝斯茨， 菲利普
更新：
要查看它是否在我打开settingsSuite的行中崩溃或退出，我在viewDidLoad的第一行添加了一个NSLog（@“Hello”）。它永远不会到达。我收到以下错误消息：
Sep 29 12:02:16 iPhone-6 kernel[0] <Notice>: xpcproxy[16389] Container: /private/var/mobile/Containers/Data/PluginKitPlugin/6A7DF264-59B2-4F38-92CB-63875B6C1469 (sandbox)
Sep 29 12:02:16 iPhone-6 Fancy Keyboard[16389] <Error>: assertion failed: 12A405: libxpc.dylib + 71820 [4BC9CA3D-4DEE-314C-ADBF-53BDCEEFE45C]: 0x7d
Sep 29 12:02:16 iPhone-6 Unknown[16389] <Error>:
Sep 29 12:02:16 iPhone-6 com.apple.xpc.launchd[1] (com.apple.xpc.launchd.domain.system) <Error>: Caller not allowed to perform action: Fancy Keyboar.16389, action = pid-local registration, code = 1: Operation not permitted, uid = 501, euid = 501, gid = 501, egid = 501, asid = 0
Sep 29 12:02:16 iPhone-6 Fancy Keyboard[16389] <Error>: _GSRegisterPurpleNamedPortInPrivateNamespace Couldn't register com.apple.accessibility.gax.client with the bootstrap server. Error: unknown error code (1100).
This generally means that another instance of this process was already running or is hung in the debugger.
Sep 29 12:02:16 iPhone-6 com.apple.xpc.launchd[1] (com.apple.xpc.launchd.domain.system) <Error>: Caller not allowed to perform action: Fancy Keyboar.16389, action = pid-local registration, code = 1: Operation not permitted, uid = 501, euid = 501, gid = 501, egid = 501, asid = 0
Sep 29 12:02:17 iPhone-6 ReportCrash[16390] <Error>: task_set_exception_ports(B07, 400, D03, 0, 0) failed with error (4: (os/kern) invalid argument)
任何想法意味着什么？