fork()的异步信号安全性 [英] Async signal safety of fork()

问题描述

根据Oracle的《多线程编程指南》 ，fork()应该在信号处理程序中可以安全使用.但是我的进程陷入了信号处理程序的内部，并跟踪以下回溯:

According to Oracle's Multithreaded Programming Guide, fork() should be safe-to-use inside signal handlers. But my process got stuck inside signal handler with to following back trace:

 #0  __lll_lock_wait_private () at   ../nptl/sysdeps/unix/sysv/linux/x86_64/lowlevellock.S:95
 #1  0x00007f86e6a9990d in _L_lock_48 () from /lib/x86_64-linux- gnu/libc.so.6
 #2  0x00007f86e6a922ec in ptmalloc_lock_all () at arena.c:242
 #3  0x00007f86e6ad5e82 in __libc_fork () at ./nptl/sysdeps/unix/sysv/linux/x86_64/../fork.c:95
 #4  0x00007f86e7d9f125 in __fork () at ./nptl/sysdeps/unix/sysv/linux/pt-fork.c:25
....
 #7  signal handler called

所以malloc在信号处理程序中不能安全使用fork怎么可能?

So as malloc is not safe to be use in signal handler how fork can be?

谢谢.

推荐答案

现在为被RedHat列为错误:

错误1422161-glibc:叉子不是异步信号安全的

...

+++该错误最初是作为错误#1422159 +++的克隆而创建的

+++ This bug was initially created as a clone of Bug #1422159 +++

POSIX要求fork是异步信号安全的.我们目前 实施不是.

POSIX requires that fork is async-signal-safe. Our current implementation is not.

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