一种方法隐含的 dup(2)在巴什输出在fd {10,11,12}一个真实用例并在之后立即关闭它，基于在此:
$ cat tags
desktop-19.9.0
foobar-1.2.3
desktop-22.9.0
mobile-24.10.0
desktop-18.9.0
desktop-21.9.0
mobile-23.10.0
foobar-1.2.4
desktop-17.8.0
desktop-20.8.0
mobile-22.9.0
desktop-16.8.0
desktop-19.8.0
mobile-21.9.0
foobar-1.2.5
desktop-15.7.0
desktop-18.7.0
mobile-20.8.0
desktop-14.7.0
desktop-17.7.0
mobile-19.8.0
desktop-13.6.0
desktop-16.6.0
mobile-18.7.0
foobar-1.2.6
desktop-12.6.0
desktop-15.6.0
mobile-17.7.0
desktop-11.5.0
desktop-14.5.0
mobile-16.6.0
desktop-10.5.0
desktop-13.5.0
mobile-15.6.0
desktop-9.4.0
desktop-12.4.0
mobile-14.5.0
desktop-8.4.0
desktop-11.4.0
mobile-13.5.0
desktop-7.3.0
foobar-1.2.7
desktop-10.3.0
mobile-12.4.0
desktop-6.3.0
desktop-9.3.0
mobile-11.4.0
desktop-5.2.0
desktop-8.2.0
mobile-10.3.0
desktop-4.2.0
desktop-7.2.0
mobile-9.3.0
desktop-3.1.0
desktop-6.1.0
mobile-8.2.0
desktop-2.1.0
desktop-5.1.0
mobile-7.2.0
desktop-1.0.0
foobar-1.2.8
desktop-4.0.0
mobile-6.1.0
cat tags | tee /dev/fd/{10,11,12} 10> >(
grep -w desktop | tail -n 3) 11> >(
grep -w mobile | tail -n 3) 12> >(
grep -w foobar | tail -n 3) 1>/dev/null
输出：
foobar-1.2.6
foobar-1.2.7
foobar-1.2.8
mobile-8.2.0
mobile-7.2.0
mobile-6.1.0
desktop-5.1.0
desktop-1.0.0
desktop-4.0.0
这是有效的，但我没有解释怎么运行的。幕后发生了什么？
为什么fd自动关闭我不清楚。可以使用显式的 open/close 来编写吗fd？
test -e /dev/fd/10 && echo open || echo close
close
编辑:
可以简化为：
cat tags | tee >(
grep -w mobile | tail -n3) >(
grep -w desktop | tail -n3) >(
grep -w foobar | tail -n3) > /dev/null
调试系统调用:
$ strace -fff -o log bash -c '
cat tags | tee >(
grep -w mobile | tail -n3) >(
grep -w desktop | tail -n3) >(
grep -w foobar | tail -n3) > /dev/null'
$ less log.*
$ grep dup log*
log.2476802:dup2(4, 1) = 1
log.2476803:dup2(3, 0) = 0
log.2476803:dup2(4, 63) = 63
log.2476803:dup2(4, 62) = 62
log.2476803:dup2(4, 61) = 61
log.2476803:dup2(3, 1) = 1
log.2476804:dup2(3, 0) = 0
log.2476805:dup2(4, 1) = 1
log.2476806:dup2(3, 0) = 0
log.2476807:dup2(3, 0) = 0
log.2476808:dup2(4, 1) = 1
log.2476809:dup2(3, 0) = 0
log.2476810:dup2(3, 0) = 0
log.2476811:dup2(4, 1) = 1
log.2476812:dup2(3, 0) = 0
$ LANG=C grep close log.*
log.2476801:close(3) = 0
log.2476801:close(4) = 0
log.2476801:close(4) = -1 EBADF (Mauvais descripteur de fichier)
log.2476801:close(3) = -1 EBADF (Mauvais descripteur de fichier)
log.2476802:close(4) = 0
log.2476802:close(3) = 0
log.2476802:close(1) = 0
log.2476802:close(2) = 0
log.2476803:close(3) = 0
log.2476803:close(4) = 0
log.2476803:close(3) = 0
log.2476803:close(4) = 0
log.2476803:close(3) = 0
log.2476803:close(4) = 0
log.2476803:close(3) = 0
log.2476803:close(3) = 0
log.2476803:close(3) = 0
log.2476803:close(3) = 0
log.2476803:close(3) = 0
log.2476803:close(3) = 0
log.2476803:close(3) = 0
log.2476803:close(3) = 0
log.2476803:close(3) = 0
log.2476803:close(4) = 0
log.2476803:close(5) = 0
log.2476803:close(0) = 0
log.2476804:close(63) = 0
log.2476804:close(4) = -1 EBADF (Mauvais descripteur de fichier)
log.2476804:close(3) = 0
log.2476804:close(3) = -1 EBADF (Mauvais descripteur de fichier)
[...]
答案1
当你写的时候tee /dev/fd/{10,11,12} 10> (…) 11> (…) 12> (…)
shell 首先创建管道，创建子进程（继承管道），关闭未使用的管道末端并重新映射文件目录。然后当命令执行时，/dev/fd/10（或11、或12）就可以打开了。
内核只显示可用的文件描述符。重定向10>的范围仅限于其命令。
如果你想strace向你展示dup2，你最好尝试strace -ff bash -c "cmd1 >(cmd2)" 2>&1 | grep dup举例。