Purpose of function export in bash

If we take a look at the following example:

# testing(){ echo hello;}
# testing
hello
# echo $(testing)
hello
# echo testing >script
# ./script
./script: line 1: testing: command not found
# source ./script
hello
# export -f testing
# ./script
hello

It tus out that a bash function needs to be exported only if you want to use it in a non-sourced script. I tried several levels of subshells, the behavior is the same. Can someone confirm this, because I find it contradictory with the claim that local variables do not exist in subshells.

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