如何在Xcode 6中的异常断点上打印异常?

时间:2015-03-02 16:51:14

标签: ios objective-c xcode

我的应用程序正在崩溃,似乎它被异常断点捕获(这是有道理的),但我无法理解崩溃的原因。

以下是我尝试的内容:

  • po $eax
  • po $rax
  • po $r0
  • po *(id *)($esp + 4)

对于上述所有尝试,我收到以下错误:

error: use of undeclared identifier '$<register name>' error: 1 errors parsing expression

我也发现了这个LLDB Command Guide,但没有找到任何有用的东西(有点令人困惑,你不知道你在寻找什么)。

如何打印坠机原因?

我正在运行iOS 8,lldb和Xcode 6。

修改

现在我明白为什么找不到这些寄存器。这是我运行register read时得到的结果:

General Purpose Registers:
x0 = 0x0000000000000001
x1 = 0x0000000000000000
x2 = 0x0000000000000000
x3 = 0x0000000195531a74  libsystem_malloc.dylib`nano_free_definite_size
x4 = 0x0000000000000000
x5 = 0x0000000000000000
x6 = 0x676e697274534643
x7 = 0x0000000000000f80
x8 = 0x00000001569d5a10
x9 = 0x0000000000000000
x10 = 0x000001a574056051
x11 = 0x0000000000000001
x12 = 0x000000000000003d
x13 = 0x0000000000000000
x14 = 0x0000000000000001
x15 = 0x0000000000000052
x16 = 0x0000000194d6e510  libobjc.A.dylib`object_setClass
x17 = 0x0000000000000000
x18 = 0x0000000000000000
x19 = 0x00000001702823f0
x20 = 0x0000000174038eaa
x21 = 0x000000019130a778  "release"
x22 = 0x0000000000000000
x23 = 0x0000000174246d20
x24 = 0x0000000174038ea0
x25 = 0x00000001895d22fa  "forwardingTargetForSelector:"
x26 = 0x00000001745186a0
x27 = 0x0000000000000000
x28 = 0x00000000a40008ff
fp = 0x0000000105757720
lr = 0x000000018462a440  CoreFoundation`___forwarding___ + 968
sp = 0x00000001057576c0
pc = 0x000000018462a440  CoreFoundation`___forwarding___ + 968
cpsr = 0x20000000

正如您所看到的,这些寄存器不包含我之前尝试过的任何寄存器。但是,我仍然无法找到例外。

xcode view

2 个答案:

答案 0 :(得分:4)

你应该输入:

po $arg1

有关详细信息,请查看:How to replace NSUncaughtExceptionHandler with definition in Breakpoint navigator?

答案 1 :(得分:0)

强调Rickster对前一个答案的评论:

确保选择正确的框架。

在调试导航器中选择最顶级的“机器代码”框后,

po $arg1为我工作。