6

以下程序的定义行为是什么(如果有)?

#include <iostream>
#include <exception>
#include <cstdlib>

void i_throw()
{
    std::cout << "i_throw()" << std::endl;
    // std::terminate() is noexcept so if the terminate handler throws...
    // then the terminate handler is called...
    // std::terminate is [[noreturn]] so don't return
    try
    {
        throw 7;
    }
    catch(...)
    {
        std::cout << "caught exception, re-throw()-ing" << std::endl;
        throw;
    }
    std::cout << "got here!" << std::endl;
    std::abort();
}

int main()
{
    std::set_terminate(i_throw);
    throw;
    std::terminate();
}

使用 gcc 和 clang 我得到以下输出:

i_throw()
caught exception, re-throw()-ing
Aborted (core dumped)

示例在前几条评论后编辑。

(我不知道为什么我同时拥有throw;std::terminate();。我不想改变这个例子,所以只是假装这两个中的一个在那里。)

4

1 回答 1

2

上面的问题可以归结为理解以下两个代码片段的行为。

示例 1: 无活动异常抛出

int main()
{
    try{
        throw;
    }catch(...){
        std::cout<<"caught"<<endl;  //we never reach here
    }
    return 0;
}

如果您运行上面的代码,它会崩溃如下

terminate called without an active exception
Aborted (core dumped)

示例 2:抛出活动异常

int main()
{
    try{
        throw 7;
    }catch(...){
        std::cout<<"caught"<<endl;  //will be caught
    }
    return 0;
}

运行它会产生可预测的输出

caught

如果生成代码的程序集 ( g++ -S option)。您会注意到以下 cxx_abi 调用 throw vs throw 7

throw;转换为call __cxa_rethrow

throw 7;转换为call __cxa_throw

这是代码__cxa_throw

extern "C" void
__cxxabiv1::__cxa_throw (void *obj, std::type_info *tinfo,
             void (_GLIBCXX_CDTOR_CALLABI *dest) (void *))
{
  PROBE2 (throw, obj, tinfo);

  __cxa_eh_globals *globals = __cxa_get_globals ();
  globals->uncaughtExceptions += 1;

  // code removed for brevity 
  //.......
  // Below code throws an exception to be caught by caller

  #ifdef _GLIBCXX_SJLJ_EXCEPTIONS
    _Unwind_SjLj_RaiseException (&header->exc.unwindHeader);
  #else
    _Unwind_RaiseException (&header->exc.unwindHeader);
  #endif

  // Some sort of unwinding error.  Note that terminate is a handler.
  __cxa_begin_catch (&header->exc.unwindHeader);
  std::terminate ();
}

因此,在 OP Codethrow 7;中会被相应的捕获并被catch(...)重新抛出throw;

这是代码__cxa__rethrow

extern "C" void
__cxxabiv1::__cxa_rethrow ()
{
  __cxa_eh_globals *globals = __cxa_get_globals ();
  __cxa_exception *header = globals->caughtExceptions; // We are not re

  globals->uncaughtExceptions += 1;

  // Watch for luser rethrowing with no active exception.
  if (header)
    {
      // Code removed for brevity
      // .....
      // Below code rethrows the exception
      #ifdef _GLIBCXX_SJLJ_EXCEPTIONS
      _Unwind_SjLj_Resume_or_Rethrow (&header->unwindHeader);
      #else
      #if defined(_LIBUNWIND_STD_ABI)
      _Unwind_RaiseException (&header->unwindHeader);
      #else
      _Unwind_Resume_or_Rethrow (&header->unwindHeader);
      #endif
      #endif
    }
  std::terminate ();
}

在这两种情况下,我们都可以看到std::terminate()尚未从__cxx_*. 被上面的 abi 抛出后,我们在代码中的以下位置。

请参阅cxx_abi以终止代码。

void
__cxxabiv1::__terminate (std::terminate_handler handler) throw ()
{
  __try 
    {
      handler ();      // Our handler has thrown an int exception
      std::abort ();
    } 
  __catch(...)  // Exception is caught here and process is aborted.
    { std::abort (); } 
}

void
std::terminate () throw()
{
  __terminate (get_terminate ());
}

概括

根据我的理解,从处理程序重新抛出异常会导致在__cxxabiv1::__terminate. 它调用的地方abort()。显然,std::terminate()[from __cxa_rethrow] 方法没有出现,这就是为什么控件从未到达 std::cout << "got here!" << std::endl;

无限递归

如果我们将 terminate_handler 更改为以下内容会发生什么:

void i_throw()
{
    std::cout << "i_throw()" << std::endl;
    throw;
    std::cout << "got here!" << std::endl;
    std::abort();
}

要理解这一点,我们可以看看__cxa_rethrow()上面提到的。

由于没有抛出活动异常,__cxa_rethrow()因此最终会调用std::terminate(),从而导致无限递归。

于 2015-02-27T10:29:36.053 回答