How to return an array in bash without using globals?

Question

I have a function that creates an array and I want to return the array to the caller:

create_array() {
  local my_list=(\"a\", \"b\", \"c\")
  echo \"${my_li         


        

Answer 1

I tried various implementations, and none preserved arrays that had elements with spaces ... because they all had to use echo.

# These implementations only work if no array items contain spaces.
use_array() {  eval echo  '(' \"\${${1}\[\@\]}\" ')';  }
use_array() {  local _array="${1}[@]"; echo '(' "${!_array}" ')';  }

Solution

Then I came across Dennis Williamson's answer. I incorporated his method into the following functions so they can a) accept an arbitrary array and b) be used to pass, duplicate and append arrays.

# Print array definition to use with assignments, for loops, etc.
#   varname: the name of an array variable.
use_array() {
    local r=$( declare -p $1 )
    r=${r#declare\ -a\ *=}
    # Strip keys so printed definition will be a simple list (like when using
    # "${array[@]}").  One side effect of having keys in the definition is 
    # that when appending arrays (i.e. `a1+=$( use_array a2 )`), values at
    # matching indices merge instead of pushing all items onto array.
    echo ${r//\[[0-9]\]=}
}
# Same as use_array() but preserves keys.
use_array_assoc() {
    local r=$( declare -p $1 )
    echo ${r#declare\ -a\ *=}
}  

Then, other functions can return an array using catchable output or indirect arguments.

# catchable output
return_array_by_printing() {
    local returnme=( "one" "two" "two and a half" )
    use_array returnme
}
eval test1=$( return_array_by_printing )

# indirect argument
return_array_to_referenced_variable() {
    local returnme=( "one" "two" "two and a half" )
    eval $1=$( use_array returnme )
}
return_array_to_referenced_variable test2

# Now both test1 and test2 are arrays with three elements