bash: notes on variable indirection

Programming_Languages/Shell/Variable_indirection.md

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+---
+categories:
+  - Programming Languages
+tags:
+  - shell
+---
+
+# Variable indirection
+
+> Write proper notes on this
+
+```sh
+function array_empty() {
+    declare -n arr=$1
+    # Proceed if array not empty:
+    if [ ${#arr[@]} -gt 0 ]; then
+        return 1 # array is not empty
+    else
+        return 0 # array is empty
+    fi
+}
+
+my_array=(1 2 3)
+
+function push() {
+    # $1 = stack name
+    local stack_name="$1"
+    if [ "${stack_name}" ]; then # stack exists
+        if array_empty "${!stack_name}"; then
+            echo "stack is empty"
+        else
+            echo "stack is not empty"
+        fi
+    else
+        echo "Error: ${stack_name} does not exist."
+    fi
+}
+
+push "my_array"
+
+```
+
+In Bash scripting, ${!stack_name} is an example of indirect variable referencing or variable indirection. It's a way to use a variable whose name is stored in another variable.
+
+Let's break it down:
+
+`${stack_name}`: This syntax is used to reference the value of a variable. So if stack_name="my_array", then ${stack_name} would return "my_array".
+
+`${!stack_name}`: Adding the ! introduces indirection. Now, instead of getting the value of stack_name, you get the value of the variable whose name is stored in stack_name. So if stack_name="my_array" and my_array=(1 2 3), then ${!stack_name} would return (1 2 3).
+
+So, in the context of your script, `${!stack_name} allows you to pass the name of an array (as a string) to your push function, and then use that string to indirectly reference the actual array within the function.