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linux - bash set -e function with double 管道

转载 作者:可可西里 更新时间:2023-11-01 11:48:46 26 4
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有人可以解释这种行为吗?我觉得很奇怪。

#!/bin/bash

set -e

_func() {
cat non-existant-file
return 0
}

_func || echo "Not seen since _func returned zero"
_func
echo "Not seen since _func returned non-zero"

输出(GNU bash,版本 4.3.42(1)-release (x86_64-pc-linux-gnu)):

$ bash func.sh 
cat: non-existant-file: No such file or directory
cat: non-existant-file: No such file or directory
$

最佳答案

如果您查看 bash 的联机帮助页,它特别提到了这一点:

特别是 如果失败的命令是 [...] 在 && 或 || 中执行的任何命令的一部分,则 shell 不会退出列出除最后的 && 或 || 之后的命令之外的内容[...]

此行为与记录/预期的完全一致。

如果你运行这样的东西,你可以自己看到:

#!/bin/bash

set -ex
# set -x will tell you exactly where the shell exits.

_func() {
cat non-existant-file
return 0
}

#_func || echo "Not seen since _func returned zero"
cat non-existant-file || _func
_func
echo "Not seen since _func returned non-zero"

来自手册页:

    set   -e      Exit immediately if a pipeline (which may consist  of  a
single simple command), a list, or a compound command
(see SHELL GRAMMAR above), exits with a non-zero sta‐
tus. The shell does not exit if the command that fails
is part of the command list immediately following a
while or until keyword, part of the test following the
if or elif reserved words, part of any command executed
in a && or || list except the command following the
final && or ||, any command in a pipeline but the last,
or if the command's return value is being inverted with
!. If a compound command other than a subshell returns
a non-zero status because a command failed while -e was
being ignored, the shell does not exit. A trap on ERR,
if set, is executed before the shell exits. This option
applies to the shell environment and each subshell envi‐
ronment separately (see COMMAND EXECUTION ENVIRONMENT
above), and may cause subshells to exit before executing
all the commands in the subshell.

这正是你这里的情况。

关于linux - bash set -e function with double 管道,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35651389/

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