Tuesday, October 4, 2022

[SOLVED] Passing arrays as parameters in bash

Issue

How can I pass an array as parameter to a bash function?

Note: After not finding an answer here on Stack Overflow, I posted my somewhat crude solution myself. It allows for only one array being passed, and it being the last element of the parameter list. Actually, it is not passing the array at all, but a list of its elements, which are re-assembled into an array by called_function(), but it worked for me. If someone knows a better way, feel free to add it here.


Solution

You can pass multiple arrays as arguments using something like this:

takes_ary_as_arg()
{
    declare -a argAry1=("${!1}")
    echo "${argAry1[@]}"

    declare -a argAry2=("${!2}")
    echo "${argAry2[@]}"
}
try_with_local_arys()
{
    # array variables could have local scope
    local descTable=(
        "sli4-iread"
        "sli4-iwrite"
        "sli3-iread"
        "sli3-iwrite"
    )
    local optsTable=(
        "--msix  --iread"
        "--msix  --iwrite"
        "--msi   --iread"
        "--msi   --iwrite"
    )
    takes_ary_as_arg descTable[@] optsTable[@]
}
try_with_local_arys

will echo:

sli4-iread sli4-iwrite sli3-iread sli3-iwrite  
--msix  --iread --msix  --iwrite --msi   --iread --msi   --iwrite

Edit/notes: (from comments below)

  • descTable and optsTable are passed as names and are expanded in the function. Thus no $ is needed when given as parameters.
  • Note that this still works even with descTable etc being defined with local, because locals are visible to the functions they call.
  • The ! in ${!1} expands the arg 1 variable.
  • declare -a just makes the indexed array explicit, it is not strictly necessary.


Answered By - Ken Bertelson
Answer Checked By - Dawn Plyler (WPSolving Volunteer)
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