I’ve written small bash scripts before, but I bit off a little more this time, and I’m trying to program a little terminal game.
Anyway, I’ve run into a weird behavior. I have a function that, among other things, writes values to an array and returns either 0 or 1.
If I write echo "${name_of_array[@]}" inside the function, I see the contents of the array printed to standard output.
If I write echo "${name_of_array[@]}" outside the function immediately after executing the function, I see the contents of the array printed to standard output. (So, clearly the array is being treated as a global variable.)
But if I write the following, regardless of which value the function returns, nothing is printed to standard output but an empty line.
if ! name_of_function ; then
echo "${name_of_array[@]}"
fi
Why is that the case? Doesn’t name_of_function get executed when evaluating the if statement? Is this some special case where all variables become local?
I realize I could just assign the function’s return value to a separate variable and then use that variable as the condition to the if statement, but it’s less elegant, and it doesn’t satisfy my curiosity. Is there any way to get the array out of the if statement alone?
It’s also possible I’m an idiot and my problem is just some random punctuation somewhere.
What is the return value ( $? ) from the function?
Not sure if
if..thenchanges variable scope: it shouldn’t.I’ve always used
[[..]]around tests, not sure if that changes variable scope.Where is the array first declared? Inside or outside the function?
Not at my computer at the moment, so I can’t give you a better answer. I’ll play with the idea later today.
The return value is either 0 or 1.
I don’t know enough about coding to know what a variable’s scope is. *SearXNG noises* You might be onto something. You think the function is being piped into a subshell?
Brackets didn’t seem to help in my script; although maybe I should make a smaller test script just to rule out incompetence. (I’m guessing on the proper syntax for that, something like
if ! [[ $(name_of_function "$arguments") ]] ; then, yes?)I’ve tried both versions.
name_of_array=()in or out of the function, alsodeclare -a -g name_of_array, if I’m remembering those arguments right, at the top of the script.Thanks for indulging my curiosity; I appreciate it.
This works fine for me:
function foo() { bar[0]="hello" bar[1]="world" return 1 } if ! foo; then echo "${bar[@]}" fihttps://onlinegdb.com/xPIFP110w
Are you getting messed up by the way bash handles exit statuses? An exit status of 0 indicates a success and a non-zero exit status indicates a failure (which allows for the different exit statuses to indicate different errors).
So
if my_func; then something; fiwill only runsomethingwhenmy_funcreturns 0. In your case, you’re using!to do the opposite so it only runs when your function returns a non-zero status.This can be quite surprising if you’re expecting the behavior found in other languages like python or C++ where 0 represents false and 1 represents true.
I’ve never used pastebin or similar services, but if you feel like wading through my mess of a coding playground, the offending code is when findclosestbox calls either collisiondetection or findbetterport, and the debug “echo” I’m interested in is the one after “This is the end of the findclosestbox function:”
https://zerobin.net/?9fa46f9871bb3607#tnEmm2JQZ3ggy2+fLp2oJwGf5r5uRcs0xj8zIvyurYQ=
password: lemmydebug
I went ahead and put the same command on either side of an if/else just to make sure I wasn’t messing up the return value. These results are interesting though; I suppose that means I just made an unrelated mistake somewhere. Incompetence is completely plausible in this case.
Edit: Wait though, now my script works, but all I did was expand the if statement into name_of_function, return_variable=$?, and if ! [[ return_variable ]]; then. I didn’t screw anything else up. So the mystery remains.