抑制信号发生

我有一个类似于此的 bash 脚本：

#!/bin/bash
timeout -s ALRM 1 sleep 5

此脚本应触发超时并返回退出代码 124。但它也会将以下文本打印到 stderr：

a.sh: line 2:   604 Alarm clock             timeout -s ALRM 1 sleep 5

bash 版本是

bash --version
GNU bash，版本 5.0.18(1)-release (arm-mut-linux-gnueabi)

是什么导致此调试输出？我如何压制它？

这是我的商店列表

# shopt
autocd          off
assoc_expand_once       off
cdable_vars     off
cdspell         off
checkhash       off
checkjobs       off
checkwinsize    on
cmdhist         on
compat31        off
compat32        off
compat40        off
compat41        off
compat42        off
compat43        off
compat44        off
complete_fullquote      on
direxpand       off
dirspell        off
dotglob         off
execfail        off
expand_aliases  on
extdebug        off
extglob         off
extquote        on
failglob        off
force_fignore   on
globasciiranges on
globstar        off
gnu_errfmt      off
histappend      off
histreedit      off
histverify      off
hostcomplete    on
huponexit       off
inherit_errexit off
interactive_comments    on
lastpipe        off
lithist         off
localvar_inherit        off
localvar_unset  off
login_shell     off
mailwarn        off
no_empty_cmd_completion off
nocaseglob      off
nocasematch     off
nullglob        off
progcomp        on
progcomp_alias  off
promptvars      on
restricted_shell        off
shift_verbose   off
sourcepath      on
xpg_echo        off

回答

[STEP 101] $ cat foo.sh
busybox timeout -s ALRM 1 sleep 5
echo rc=$?
/usr/bin/timeout -s ALRM 1 sleep 5
echo rc=$?
[STEP 102] $ bash foo.sh
foo.sh: line 1: 86649 Alarm clock             busybox timeout -s ALRM 1 sleep 5
rc=142
rc=124
[STEP 103] $

{ timeout ...; } 2> /dev/null

{ timeout ...; } 2> /dev/null