c++ - valgrind 未处理的指令字节 : 0xF 0xB 0xFF 0x85

Question

我尝试在 valgrind 3.6.1 下运行使用 Apple GCC 3.2.1(强制 32 位模式，仅限 x86)编译的程序，但我得到了初始化阶段出现以下错误:

vex x86->IR: unhandled instruction bytes: 0xF 0xB 0xFF 0x85
==80746== valgrind: Unrecognised instruction at address 0x2a6c2a9.
==80746== Your program just tried to execute an instruction that Valgrind
==80746== did not recognise.  There are two possible reasons for this.
==80746== 1. Your program has a bug and erroneously jumped to a non-code
==80746==    location.  If you are running Memcheck and you just saw a
==80746==    warning about a bad jump, it's probably your program's fault.
==80746== 2. The instruction is legitimate but Valgrind doesn't handle it,
==80746==    i.e. it's Valgrind's fault.  If you think this is the case or
==80746==    you are not sure, please let us know and we'll try to fix it.
==80746== Either way, Valgrind will now raise a SIGILL signal which will
==80746== probably kill your program.
==80746== 
==80746== Process terminating with default action of signal 4 (SIGILL)
==80746==  Illegal opcode at address 0x2A6C2A9

你能告诉我这个指令是什么吗？我应该怎么做？如果我在 gdb 下运行我的应用程序，我可以毫无问题地通过此代码区域...

Answer 1

字节序列0xF 0xB是操作码UD2。

这是一个已定义的“未定义指令”，如果它有意义的话:有许多可能的操作码是不合法的，但这个是专门保留的指令，保证引发 #UD 无效的操作码异常，即使在未来的处理器上也是如此。

有一个(我只能想到一个)隐约合理的原因可以解释为什么它可能被代码故意执行:GCC 内置的__builtin_trap() 生成x86 上的 UD2 指令，我偶尔会看到它代替 abort() 来导致 fatal error ，该错误将被调试器捕获。