Does Bash have private stack frames for recursive function calls?

Question

Consider the following code:

recursion_test() {
    local_var=$1
    echo "Local variable before nested call: $local_var"

    if [[ $local_var == "yellow" ]]; then
        return
    fi

    recursion_test "yellow"

    echo "Local variable changed by nested call: $local_var"
}

Output:

Local variable before nested call: red
Local variable before nested call: yellow
Local variable changed by nested call: yellow

In other programming languages such as Java each method invocation has a separate private stack frame on which local variables are kept. So nested invocation of a method can't modify variables in the parent invocation.

In Bash do all invocations share the same stack frame? Is there a way to have separate local variables for different invocations? If not, is there a workaround to write recursive functions properly so that one invocation does not affect the other?

Answer 1

You want the local builtin. Try local local_var=$1 in your example.

NOTE: You still have to be careful as local isn't completely private as in a C stack variable. It's more like javascript's var which means any called [child] function can get at it [vs. js's let, which is totally private].

Here's an example:

recursion_test() {
    local local_var=$1
    echo "Local variable before nested call: $local_var"

    # NOTE: if we use other_func_naughty instead, we get infinite recursion
    other_func_nice

    if [[ $local_var == "yellow" ]]; then
        return
    fi

    recursion_test "yellow"

    echo "Local variable changed by nested call: $local_var"
}

other_func_naughty() {
    local_var="blue"
}

other_func_nice() {
    local local_var="blue"
}

recursion_test red

So, if you use local <whatever>, just be sure to be consistent (i.e. all functions declare it the same way)