Bash function name clash with system binary

bashshell-script

In my scripts I have a function called messages. I wrote it in Linux Mint, with no problems running it, and when I moved it to a Debian Buster station, the function clashes with /usr/bin/messages.

I have a startup script that calls the script messages:

startup_script

# call to messages script
. messages

messages

messages() {
  # reformat the arguments and return them
}

later on startup_script

messages "This is a message"

Which throws

./startup_script: line 35: .: /usr/bin/messages: cannot execute binary file
messages: could not open mailbox `/path/to/my/script/<string passed to my function>': No such file or directory

So I get a bunch of errors related to /usr/bin/messages being called instead of my function.

After adding type messages "This is a message", the relevant output is:

messages is /usr/bin/messages

I have the option of renaming my function¹, but maybe there's a better way to handle this situation.

How do I tell my script to ignore system binaries and use my own functions?

^{¹ The function is called in several scripts, many times, so it is not the easiest option to just change the name.}

Best Answer

This is how . file works:

If file does not contain a <slash>, the shell shall use the search path specified by PATH to find the directory containing file.

This behavior is specified by POSIX.

Your first error is

./startup_script: line 35: .: /usr/bin/messages: cannot execute binary file

It's similar when you call . echo:

-bash: .: /bin/echo: cannot execute binary file

You're trying to source the binary file /usr/bin/messages. In effect your file where the function is defined is not sourced at all, the function is not defined in the current script. This means later messages still means /usr/bin/messages, not the function.

The relevant line should be . ./messages or . /full/path/to/messages (i.e. path to your file, not to the binary).

Related Solutions

Bash – automatically load shell scripts from /usr/bin

The alias you have defined will only take effect once the alias command is actually called. Thus if you were to do this:

$ source myscript
$ e "Hello"

It should work. However, this is clearly not ideal since you want e to be available whenever you start bash. Luckily, bash provides a number of ways to automatically run commands when the shell is started. For the full details of how bash starts up, see the manual page. For our purposes, it is enough to know that ~/.bashrc (that is, a file named ".bashrc" in your home directory) is run as your shell starts up. One way to make your alias available at startup would be to add this line at the end of ~/.bashrc:

source myscript

However, if you were to do this for every alias you wanted, your /usr/bin folder would likely become a mess. And, if you are on a multi-user system, filling /usr/bin/ with scripts like this may cause other users problems as well. Thus, it is better to place your aliases right inside .bashrc and forgo the separate script all together. Since you are using Ubuntu, inside your .bashrc file you probably have something that looks like this:

if [ -f ~/.bash_aliases ]; then
. ~/.bash_aliases 
fi

This code looks for a file call .bash_aliases in your home directory and runs anything it finds in that file as well. If you have this, or if you add this code to your .bashrc, you could also put your alias in ~/.bash_aliases. This provides an easy way to keep all your aliases in one place and keep your .bashrc file uncluttered.

Bash – Run a script on multiple shells

It would probably be easier to have two separate versions of the script. Since it is short it may not be worth adding extra code to handle the differences between the two formats as jw013 suggested. On the other hand if you had a larger script it would probably be easier to have one and make the script execute different commands depending on where it is running from.

Best Answer

Related Solutions

Bash – automatically load shell scripts from /usr/bin

Bash – Run a script on multiple shells

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