Bash: find functions to execute before files?

dedec0 · 11-18-2015, 07:27 AM

I have something like this in my bashrc, already sourced:

Code:

# Make something...
function dir() 
{ 
    if (($# > 0)); then 
        nome="aaa/$1"; 
    else 
        nome='aaa/bbb' 
    fi 
    echo "my dir" 
    # some
    # more
    # commands
}

But when I call dir from my shell, I get the output of the dir program in Ubuntu 14.04.

I have installed in my user folder some programs. And I changed the PATH parts to make my programs be found first, if any with a same name also exists in the machine (this is just the order each directory gets in the final $PATH value).

Now I want my functions to be executed instead of any programs with the same names. How can I do it?

(renaming my function does not count! :P hehehe)

translator1111 · 11-18-2015, 07:56 AM

Dear dedec0,
please tell us how do you call dir from your shell.
Are you sure you are calling your function dir or instead are you calling the program dir?

Regards,
M.

berndbausch · 11-18-2015, 08:00 AM

What makes you certain it's sourced?
To be 100 percent sure, echo something before and after the function definition.
Also, when you type set, do you see your function?

rtmistler · 11-18-2015, 12:06 PM

Code:

$ which dir
$ alias

Are both good things to check as well here.

Habitual · 11-18-2015, 01:07 PM

Quote:

Originally Posted by rtmistler

Code:

$ which dir
$ alias

Code:

type dir

too!

dedec0 · 11-21-2015, 07:26 AM

Quote:

please tell us how do you call dir from your shell.
Are you sure you are calling your function dir or instead are you calling the program dir?

That echo "my dir" was put here to show that my function is not being called. What is being called is:

Code:

$ whereis dir
dir: /bin/dir /usr/share/man/man1/dir.1.gz

-------------------------------

Quote:

What makes you certain it's sourced?
To be 100 percent sure, echo something before and after the function definition.
Also, when you type set, do you see your function?

Before making this thread I opened a new session (calling bash, opening another terminal window, anything that will surely source my bashrc without side effects from any existing environment). But to answer you, set|grep -e '^dir ()' -A 9 outputs:

Code:

$ set | grep -e '^dir ()' -A 9
5390:dir () 
5391-{ 
5392-    [
5393-    # my function! :D
5394-    ]
5399-}

-------------------------------

Quote:

Code:

$ which dir
$ alias

Are both good things to check as well here.

Code:

$ which dir
/bin/dir
$ alias | grep dir
dir='...'

I had an alias for dir... :P

Code:

$ # after fixing it...
$ type dir
dir is a function
dir () 
{ 
    [
    # my function! :D
    ]
}

'type' is a nice and new thing for me.

Thank you all!