Linux 为什么用 grep -q 退出代码 141？
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Why exit code 141 with grep -q?
提问by Sandra Schlichting
Can someone explain why I get exit code 141 from the below?
有人可以解释为什么我从下面得到退出代码 141 吗？
#!/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[@]}
I get
我得到
...
a 0 0
b 141 0
c 0 0
From my understanding exit code 141 is a failure, but the line above gives zero, so it should be success, I would say.
根据我的理解，退出代码 141 是失败的，但上面的行给出了零，所以我会说它应该是成功的。
采纳答案by dogbane
This is because grep -qexits immediately with a zero status as soon as a match is found. The zfscommand is still writing to the pipe, but there is no reader (because grephas exited), so it is sent a SIGPIPEsignal from the kernel and it exits with a status of 141.
这是因为grep -q一旦找到匹配项，就会立即以零状态退出。该zfs命令仍在写入管道，但没有读取器（因为grep已退出），因此它SIGPIPE从内核发送一个信号，并以141.
Another common place where you see this behaviour is with head. e.g.
您看到此行为的另一个常见位置是使用head. 例如
$ seq 1 10000 | head -1
1
$ echo ${PIPESTATUS[@]}
141 0
In this case, headread the first line and terminated which generated a SIGPIPEsignal and seqexited with 141.
在这种情况下，head读取第一行并终止它生成一个SIGPIPE信号并seq退出141。
See "The Infamous SIGPIPE Signal" from The Linux Programmer's Guide.
请参阅Linux 程序员指南中的“臭名昭著的 SIGPIPE 信号”。
回答by volferine
I'm not familiar with zfs list, but I guess it complains about its standard output being closed - grep -qexits immediately when a match is found, unlike grep.
我不熟悉zfs list，但我猜它抱怨它的标准输出被关闭 -grep -q当找到匹配时立即退出，与grep.
回答by Elizandro - SparcBR
Another option would be to not use a pipe, but use a process substitution:
另一种选择是不使用管道，而是使用进程替换：
grep -q tank <(zfs list)
grep -q tank <(zfs 列表)
Update: I guess is the same thing, as the process run inside parentheses will also receive sigpipe.
更新：我猜是同样的事情，因为在括号内运行的进程也会收到 sigpipe。
