这是一个 Zshell 问题,尽管 Bash 如果它具有 >(command) 语法(即此类进程替换),也可以提示解决方案。这个非常基本的代码解释了所有内容:
% fun() {
setopt localtraps
trap "echo waiting >> /tmp/out; sleep 2; echo bye >> /tmp/out; exit 1;" EXIT
echo "My PID: $sysparams[pid]" # needs zsh/system module
repeat 1000; do
read -r -t 1 line
done
}
% exec {MYFD}> >(fun)
% exit
以上工作 - fun() 将接收陷阱,两条消息将出现在 /tmp/out 中,“exit 1”将关闭进程。
我的问题:“EXIT”可以用一些实际信号代替吗?我尝试过 PIPE、HUP、INT、TERM,但都不起作用。
答案1
你的代码并不能解释一切。我不知道你想做什么。不过,我可以回答你标题中的问题:>(…)
当主 shell 退出时,进程没有收到信号。它退出是因为它到达了脚本的末尾,此时它运行EXIT
陷阱直到执行exit
内置函数。
如果您认为脚本会被提前终止,因为您认为每个read -t 1
调用将花费一秒钟:不,它们不会,它们会在父进程退出后立即返回。当父进程退出时,子read
shell 中的调用会尝试从关闭的管道中读取数据,并且底层read
系统调用会立即返回,但没有可用数据。
答案2
bash
根据和的手册zsh
,是的,您确实可以发出trap
任何信号。
bash
:
trap [-lp] [[arg] sigspec ...]
The command arg is to be read and executed when the shell receives signal(s) sigspec. If arg is absent (and there is a single sigspec) or -, each specified signal is
reset to its original disposition (the value it had upon entrance to the shell). If arg is the null string the signal specified by each sigspec is ignored by the shell
and by the commands it invokes. If arg is not present and -p has been supplied, then the trap commands associated with each sigspec are displayed. If no arguments are
supplied or if only -p is given, trap prints the list of commands associated with each signal. The -l option causes the shell to print a list of signal names and their
corresponding numbers. Each sigspec is either a signal name defined in <signal.h>, or a signal number. Signal names are case insensitive and the SIG prefix is optional.
zsh
:
trap [ arg ] [ sig ... ]
arg is a series of commands (usually quoted to protect it from immediate evaluation by the shell) to be read and executed when the shell receives any of the signals speci-
fied by one or more sig args. Each sig can be given as a number, or as the name of a signal either with or without the string SIG in front (e.g. 1, HUP, and SIGHUP are
all the same signal).