How to pass an argument passed to a bashrc function?

Asked Jul 29 '21 at 20:24

Active Jul 29 '21 at 20:27

Viewed 31 times

I added a custom command to bash and can't pass arguments into it

I have a bnpmf.sh file in /usr/bin which specifies the bnpmf function

Now I source the bnpmf.sh file from /etc/bash.bashrc with source /usr/bin/bnpmf.sh

When I iterate over the arguments in the function, by arguments I mean shell: bnpmf arg1 arg2 ,they simply do not show up whatsoever, which is incredibly weird.

I use while [[ $@ -gt 0 ]] do to iterate over the arguments

How do I get these arguments recognized in my function

edited Jul 29 '21 at 20:27

asked Jul 29 '21 at 20:24

TheEyeOfAr3s

Function arguments are in `$1`, `$2`, etc. just like script arguments. – Barmar Jul 29 '21 at 20:26
The number of arguments is `$#`, not `$@`. – Barmar Jul 29 '21 at 20:27
Any shell scripting tutorial should explain all of this. – Barmar Jul 29 '21 at 20:28
You need to post your code. Also, check it first at shellcheck.net. – Barmar Jul 29 '21 at 20:28
To iterate over the arguments, you use `for argument in "$@"; do ...`. Inside the loop body, the respective argument is then stored in the variable `argument`. – user1934428 Jul 30 '21 at 07:16

How to pass an argument passed to a bashrc function?

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