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Asked: 2018-01-09 15:31:14 +0800 CST

How can I apply "shopt -s extdebug" to my shell by default?

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I am trying to build a pre-exec checker in bash (using bash-preexec.sh)

In order to fail the command when the pre-exec hook fails, I need to run shopt -s extdebug (which fails the command when trap DEBUG command returns non-zero RC).

If I run shopt -s extdebug in my shell as a command, this works perfectly fine (the preexec checker just grepped for "OK" in a file containing word "FAIL", so command shouldn't execute):

> grep shopt .bashrc
> bash
bash-4.1$ echo TEST2
TEST2   <=== Expected, since I didn't turn on extdebug yet
bash-4.1$ shopt -s extdebug > /dev/null 2>&1
bash-4.1$ echo TEST2
bash-4.1$ # WHAT WAS EXPECTED. Command failed to execute.

However, when I add the command to my .bashrc, it has no effect:

> grep shopt .bashrc
shopt -s extdebug > /dev/null 2>&1
> bash
bash: /usr/share/bashdb/bashdb-main.inc: No such file or directory
bash: warning: cannot start debugger; debugging mode disabled
bash-4.1$ echo TEST2
TEST2

What am I doing wrong, and how can I execute this in .bashrc (I need to execute it anytime the user logs in)?

bash

1 Answers

  1. OK, figured out the answer.

        set -o functrace > /dev/null 2>&1

    is required (in addition to shopt -s extdebug) to have the functionality work in sub-shells. And somehow, .bashrc is treated as supershell of actual interactive shell, in a sense, if you don't add this, it won't take effect on interactive command prompt.

    So, you need to add both command to .bashrc

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