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Pulling variables from functions executed as conditions (programming.dev)

submitted 7 months ago by Ballig@programming.dev to c/shell@programming.dev

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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.

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