问题描述
有人能解释为什么我得到的退出code 141从下面？
Can someone explain why I get exit code 141 from the below?
#!/usr/bin/bash
set -o pipefail
zfs list | grep tank
echo a ${PIPESTATUS[@]}
zfs list | grep -q tank
echo b ${PIPESTATUS[@]}
cat /etc/passwd | grep -q root
echo c ${PIPESTATUS[@]}
我得到
...
a 0 0
b 141 0
c 0 0
从我的理解退出code 141是失败的，但上面的行给出零，所以它应该是成功的，我会说。
From my understanding exit code 141 is a failure, but the line above gives zero, so it should be success, I would say.
推荐答案
这是因为的grep -q 立即用零状态，一旦发现匹配退出。在 ZFS 命令仍然是写入管道，但没有读者（因为的grep 已退出），所以它从内核发送 SIGPIPE 信号，并将其与的状态退出 141 。
This is because grep -q exits immediately with a zero status as soon as a match is found. The zfs command is still writing to the pipe, but there is no reader (because grep has exited), so it is sent a SIGPIPE signal from the kernel and it exits with a status of 141.
如果您看到这个行为的另一个共同的地方是头。例如。
Another common place where you see this behaviour is with head. e.g.
$ seq 1 10000 | head -1
1
$ echo ${PIPESTATUS[@]}
141 0
在这种情况下，头读的第一线，终止其产生的 SIGPIPE 信号和 SEQ 与 141 退出。
In this case, head read the first line and terminated which generated a SIGPIPE signal and seq exited with 141.
请参阅从Linux的程序员指南。
See "The Infamous SIGPIPE Signal" from The Linux Programmer's Guide.
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