How to get caller function call method in bash

Question

Based on the following code below, is it possible to obtain the value of $caller_method from the pseudocode function below whether a function's caller has made a function call normally eg: mytest 1 or using subshell style eg: echo "(mytest 1)".

#!/bin/bash
function mytest() {
  # THIS IS PSEUDOCODE
  if $caller_method=directly; then
     echo "THIS WAS CALLED DIRECTLY"
     # Do other stuff
  elif $caller_method=inside_a_subshell; then
     echo "THIS WAS CALLED INSIDE A SUBSHELL"
     # Do other stuff
  fi
 # END OF PSEUDOCODE

}
# CALLER 
# Calling mytest directly
mytest 1
# Calling mytest inside a subshell
echo "$(mytest 1)"

expected output:

THIS WAS CALLED DIRECTLY
THIS WAS CALLED INSIDE A SUBSHELL

So, does the mytest() function able to understand or store an information whether it has been called using this method mytest 1 or $(mytest 1) ?

Also, I don't want to have any extra arguments passed from the caller function such as $(mytest 1 call_inside_a_subshell) or mytest 1 call_directly

Answer 1

I just found out this issue is related how to detect if we are in a subshell using a built-in variable $BASHPID How can I detect if I'm in a subshell?

So the code can be written:

#!/bin/bash
function mytest()
{
    if [ "$$" -eq "$BASHPID" ]; then
        echo "THIS WAS CALLED DIRECTLY"
    else
        echo "THIS WAS CALLED INSIDE A SUBSHELL"
    fi
}
Calling mytest directly
mytest 1
Calling mytest inside a subshell
echo "$(mytest 1)"

Answer 2

I think it can be done. Try this:

#!/bin/bash
function mytest() {
    # THIS IS PSEUDOCODE
    if [ $ORIGINALBASHPID -eq $BASHPID ]; then
        echo "THIS WAS CALLED DIRECTLY"
        # Do other stuff
    else
        echo "THIS WAS CALLED INSIDE A SUBSHELL"
        # Do other stuff
    fi
    # END OF PSEUDOCODE
}
CALLER
ORIGINALBASHPID=$BASHPID
Calling mytest directly
mytest 1
Calling mytest inside a subshell
echo "$(mytest 1)"

It outputs:

THIS WAS CALLED DIRECTLY
THIS WAS CALLED INSIDE A SUBSHELL