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