Edit: SOLVED. Thank you all for your incredible insights! All of you helped me improve my code and knowledge! Special thanks to @Quibblekrust@thelemmy.club who just NAILED it. :)

I’m playing around with Bash just to learn.

LIST=$(ls); for i in $LIST; do echo "I found one!"; done

The variable “i” could literally be anything, as long as it doesn’t have a special meaning for Bash, in which case I’d have to escape it, right? Anyway, my real question is: how does do (or rather the whole for-expression) know that “i” here means “for every line/item that ls outputs”? The above one liner works great and writes “I found one!” the number of times corresponding to the number of lines or items that ls outputs. But I would like to understand why it worked…

I’m a complete beginner at both Bash and C, but I understand some basic concepts.

  • harsh3466@lemmy.ml
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    4 days ago

    You’ve got a few things going on to be broken down here.

    And forgive me if anything I say here seems condescending, it’s not meant to be, I just like to be very explanatory with things like this and to assume the reader may not know anything about anything. (Not as an insult, but simply as a gap in knowledge).

    Also, while I’m proficient at Bash, I’m no expert.

    LIST=$(ls): Here you’ve stored the output of the ls command to the variable LIST, which gives you a list of items in the given directory, in this case, whichever directory the command is run from. It’s also a good idea to quote the variable assignment like this: "$(ls)".

    for i in $LIST;: This is the first part of the for loop statement, which is an iterator, meaning, it will loop or iterate over every item in the given variable/parameter/group of iterable items.

    The i here, as you said could be anything. You could say for file in $LIST; or for item in $LIST;. It doesn’t matter, because it’s just a variable name that you are using in the first part of the for statement.

    So what bash will do with this is loop over the list, and take each item in the list, and assign it to the variable i, which will allow you do act upon that single item by calling the variable i in some other commands.

    do echo "I found one!";: This is the next part of the for loop, which is the instruction set to be executed inside the for loop. Here is where you can act upon the items in your list that have been assigned to the variable i.

    In your case, you’re just printing a statement to stdout (standard out), that says, “I found one!”

    It’s like saying, for each item in this list, print “I found one!”

    So if there are 20 items in the list, it will print that statement 20 times.

    However, maybe you want to print the item itself as part of a statement. So instead of “I found one!”, you could do something like:

    do echo "I found $i!"

    Which then would print “I found some-filename-or-directory-here!” for each item in your list.

    done: Finally, the done statement tells bash that this is the end of the for loop. So any commands after the done statement will only run once the for loop has iterated over all items in the list and executed the commands inside the for loop for each item on the list.

    A couple of notes:

    The ; is used as a command separator or terminator. So bash knows to first run LIST=$(ls) before it attempts to run whatever the next command might be.

    In bash, it’s good practice to always quote your variables like so: for i in "$LIST";. This is to avoid errors for characters that might need escaping like whitespace, backslashes, and other special characters.

    With that in mind, if you’re running a command like echo "I found $i!", you don’t need to quote the variable again, because it’s already inside a quote set.

    Further, it’s not absolutely necessary, but it can also be a good idea to also enclose all of your variables in {}, so whenever you use a variable, you’d do something like: "${LIST}"

    This not only more clearly identifies variables in your bash scripts/commands, but is necessary when using bash’s parameter expansion, which is pretty great.

    • emotional_soup_88@programming.devOP
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      4 days ago

      I was a teacher for some years and I absolutely understand your style of explanation. I don’t find it condescending at all! Thank you so much for the in depth guidance! Some of it I already knew, some of it I didn’t. Anyhow, a new perspective is always appreciated! :) God, Bash (GNU/Linux in general) is so much fun!

      • harsh3466@lemmy.ml
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        4 days ago

        I was also a teach for a number of years! Hello fellow teacher. :)

        I agree. Bash, and GNU/Linux in general is amazing. My recent foray has been into Python, and I’m having an utter blast writing code and learning.

    • rycee@lemmy.world
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      4 days ago

      Wouldn’t for i in "$LIST"; just result in a single loop iteration with $i being the entirety of $LIST?

      • harsh3466@lemmy.ml
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        4 days ago

        It would not, as @Quibblekrust@thelemmy.club explained in their comment (which I neglected to include in my explanation), Bash uses a special variable called IFS when executing for loops like this. IFS stands for Input Field Separators, and is a list of one of each type of whitespace (tab, space, and newline), and uses these as separators automatically.

        So instead of taking that whole ls output as one string of text, the for loop automatically separates it into an iterable list of strings using the newline separator.

        • rycee@lemmy.world
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          4 days ago

          I’m pretty sure that IFS does not apply to quoted strings since word splitting happens before the quote removal (see Shell Expansion).

          $ ( files=$(ls); IFS=$'\n' ; for x in $files; do echo $x; done )
          file a.txt
          file b.txt
          plainfile.txt
          
          $ ( files=$(ls); IFS=$'\n' ; for x in "$files"; do echo $x; done )
          file a.txt file b.txt plainfile.txt
          
        • frongt@lemmy.zip
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          4 days ago

          Which makes it real fun when you have spaces in filenames!

          Really you shouldn’t use ls as input to for. Use find -exec or something.