为什么 _exit 会失败？

Question

我正在寻找 C 中 abort() 函数的源代码，我遇到了这个：

abort.c:

/* Copyright (C) 1991-2019 Free Software Foundation, Inc.
   This file is part of the GNU C Library.
   The GNU C Library is free software; you can redistribute it and/or
   modify it under the terms of the GNU Lesser General Public
   License as published by the Free Software Foundation; either
   version 2.1 of the License, or (at your option) any later version.
   The GNU C Library is distributed in the hope that it will be useful,
   but WITHOUT ANY WARRANTY; without even the implied warranty of
   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
   Lesser General Public License for more details.
   You should have received a copy of the GNU Lesser General Public
   License along with the GNU C Library; if not, see
   <http://www.gnu.org/licenses/>.  */
#include <libc-lock.h>
#include <signal.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sigsetops.h>
/* Try to get a machine dependent instruction which will make the
   program crash.  This is used in case everything else fails.  */
#include <abort-instr.h>
#ifndef ABORT_INSTRUCTION
/* No such instruction is available.  */
# define ABORT_INSTRUCTION
#endif
/* Exported variable to locate abort message in core files etc.  */
struct abort_msg_s *__abort_msg __attribute__ ((nocommon));
libc_hidden_def (__abort_msg)
/* We must avoid to run in circles.  Therefore we remember how far we
   already got.  */
static int stage;
/* We should be prepared for multiple threads trying to run abort.  */
__libc_lock_define_initialized_recursive (static, lock);
/* Cause an abnormal program termination with core-dump.  */
void
abort (void)
{
  struct sigaction act;
  sigset_t sigs;
  /* First acquire the lock.  */
  __libc_lock_lock_recursive (lock);
  /* Now it's for sure we are alone.  But recursive calls are possible.  */
  /* Unblock SIGABRT.  */
  if (stage == 0)
    {
      ++stage;
      __sigemptyset (&sigs);
      __sigaddset (&sigs, SIGABRT);
      __sigprocmask (SIG_UNBLOCK, &sigs, 0);
    }
  /* Send signal which possibly calls a user handler.  */
  if (stage == 1)
    {
      /* This stage is special: we must allow repeated calls of
         `abort' when a user defined handler for SIGABRT is installed.
         This is risky since the `raise' implementation might also
         fail but I don't see another possibility.  */
      int save_stage = stage;
      stage = 0;
      __libc_lock_unlock_recursive (lock);
      raise (SIGABRT);
      __libc_lock_lock_recursive (lock);
      stage = save_stage + 1;
    }
  /* There was a handler installed.  Now remove it.  */
  if (stage == 2)
    {
      ++stage;
      memset (&act, '[=10=]', sizeof (struct sigaction));
      act.sa_handler = SIG_DFL;
      __sigfillset (&act.sa_mask);
      act.sa_flags = 0;
      __sigaction (SIGABRT, &act, NULL);
    }
  /* Try again.  */
  if (stage == 3)
    {
      ++stage;
      raise (SIGABRT);
    }
  /* Now try to abort using the system specific command.  */
  if (stage == 4)
    {
      ++stage;
      ABORT_INSTRUCTION;
    }
  /* If we can't signal ourselves and the abort instruction failed, exit.  */
  if (stage == 5)
    {
      ++stage;
      _exit (127);
    }
  /* If even this fails try to use the provided instruction to crash
     or otherwise make sure we never return.  */
  while (1)
    /* Try for ever and ever.  */
    ABORT_INSTRUCTION;
}
libc_hidden_def (abort)

翻看，看到这个，很疑惑：

调用_exit后有：

If even this fails try to use the provided instruction to crash or otherwise make sure we never return.

为什么 _exit 会失败？ _exit 永远不会 return 并终止程序。为什么 _exit 无法终止程序？

Answer 1

有时在内存已损坏的情况下调用 abort()。这可能会导致 _exit() 以某种方式失败。

或者 OS 中可能存在错误，导致 _exit() 无法正常工作。

或者该程序已链接到替换 _exit().

的库

所有这些事情都不太可能（并且在内存损坏的情况下，它很容易导致 abort() 本身失败），但 abort() 正试图涵盖所有基础。

Answer 2

在许多系统上，会有一些代码运行ning 本身不在“进程”中。举个简单的例子，负责在进程之间执行任务切换的代码通常无法在旧进程或新进程中运行，而是在与任何一个都不关联的上下文中。尽管在此类上下文中只应调用少量系统函数，exit() 和 abort() 都不在其中，但不应该发生的事实并不能保证它不会发生。有时可能会出现这样的情况，即所有行动方案都被认为是糟糕的，但调用 abort 可能被视为其中最不糟糕的。

如果在代码处于与任何可识别进程无关的上下文中时调用 exit()，它将无法执行其正常职责。它的调用者不太可能为它可能 return 的可能性做好准备，因此其最合乎逻辑的行动过程可能是调用 abort()。另一方面，由于 abort() 最自然的做法是调用 exit() 来清理当前进程，这会产生致命的递归相互依赖。为了适应这一点，abort 函数有一个专用的静态对象，用于确定它是否被递归调用，如果是，则尝试不同的方法强制程序退出。 exit() 函数很可能也采用了类似的措施，因此如果某些清理操作触发 abort() 然后调用 exit()，它将能够尝试执行清理的任何部分跟随失败的部分。

请注意，如果正常的中止方法失败，让中止函数进入一个具有虚拟副作用的无限循环可能与任何其他操作过程一样好，但在某些情况下，无限循环会反复尝试execute ABORT_INSTRUCTION 可能更好，因为其他监督子系统可能能够识别执行上下文何时卡在除了 ABORT_INSTRUCTION 之外什么都不做的循环中并采取适当的行动（例如强制系统重启，取决于崩溃的代码一直在做什么）。

为什么 _exit 会失败？

Why would _exit fail?

c

abort