bash:在调试陷阱中保留$ _ [英] bash: preserve $_ in a DEBUG trap

问题描述

在 bash 中，手册上说 $ _ 应该是最后执行的命令的最后一个参数".

In bash, the manual says $_ should be "the last argument of the last command executed".

这是预期的行为:

$ echo 1 2 3
1 2 3
$ echo $_
3
$ 

在使用 DEBUG 陷阱时， $ _ 是基于陷阱运行的最后一条命令(在以下示例中为 date )比用户上次输入的命令要高:

When using a DEBUG trap, $_ is based upon the last command run by the trap (date in the example below) rather than the last command the user entered:

$ exec bash
$ unset PROMPT_COMMAND
$ trap date DEBUG
$ echo 1 2 3
Sat Dec  3 11:54:33 ICT 2016
1 2 3
$ echo $_
Sat Dec  3 11:54:37 ICT 2016
date
$ # User blinks in confusion and says "but I didn't just type date!"

似乎默认情况下，设置 DEBUG 陷阱会更改 $ _ 的默认行为.

It seems that by default, setting a DEBUG trap changes the default behaviour of $_.

chruby 和有趣的是，这在 zsh 中可以正常工作:

Interestingly, this works as expected in zsh:

$ zsh
genesis% trap date DEBUG
genesis% echo 1 2 3
Sat Dec  3 12:18:01 ICT 2016
1 2 3
genesis% echo $_
Sat Dec  3 12:18:04 ICT 2016
3
genesis%

使用 DEBUG 陷阱时如何保存 $ _ 的值?

How can I preserve the value of $_ when using a DEBUG trap?

推荐答案

值得注意的是，执行的最后一个命令的最后一个参数"包括文字文本执行的最后一个命令"，而不是用户输入的最后一个命令";bash的行为正如其文档在这方面所承诺的那样.

It's worth noting that "the last argument of the last command executed" includes the literal text "last command executed", not "last command entered by the user"; bash is behaving as its documentation promises in this respect.

但是请不要担心:除非您的陷阱曾经返回非零值(因此中止了它们之前运行的命令)，否则就很容易解决:

But never mind that: Unless your traps are ever returning nonzero values (and thus aborting commands they run before), this is easy enough to work around:

trapfunc() { local old_=$1; date; : "$old_"; }
trap 'trapfunc "$_"' DEBUG

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