为什么要退出code 141使用grep -q？ [英] Why exit code 141 with grep -q?

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.

请参阅臭名昭著的SIGPIPE信号从Linux的程序员指南。

See "The Infamous SIGPIPE Signal" from The Linux Programmer's Guide.

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