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c - 为什么 _exit 会失败？

问题描述

我正在寻找abort()C 中函数的源代码，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, '\0', 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)

翻阅了一下，我看到了这个并且非常困惑：

调用后_exit有：

如果即使这失败了，请尝试使用提供的指令崩溃或以其他方式确保我们永远不会返回。

为什么会_exit失败？_exit永远不会返回并终止程序。为什么地球上将_exit无法终止程序？

标签： cabort

解决方案

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

或者操作系统中可能存在阻止_exit()工作的错误。

或者该程序与替换_exit().

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

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