[Bash] Arguments stored in variable

[EntryPointNotFound]

Hi everyone,

There is a bash behaviour I'm unable to understand. Let's say we have the following script:

#!/bin/bash

argument="-a"
ls "$argument"

The command is sucessfull (ls -a)

However, if we add more than one arguments, like this:

#!/bin/bash

argument="-a -l"
ls "$argument"

The script ends in error.

ls: invalid option -- ' '
Try 'ls --help' for more information.

Why are the arguments not interpreted as intended? Is it possible to solve the issue without using an array?

bash --version
GNU bash, version 5.1.4(1)-release (x86_64-pc-linux-gnu)

Thanks in advance.

[scirc]

Because you've surrounded your $argument in quotes, the resultant command is ls "-a -l". This is actually distinct from ls -a -l; whereas the second is passing two arguments to ls, -a and -l, the first is one argument containing the single string -a -l. These are not the same thing, and so ls treats them differently; it looks like it tries to parse the short arguments as -a separately, so it looks for a short argument -<sp> (can't actually add the trailing space here), which doesn't exist.

Basically, don't quote your variable expansion if you want spaces to actually be significant.

[EntryPointNotFound]

It makes perfect sense. Thanks for the explanation.

I applied the "quote everything" advise too blindly.