Because the first fork creates two processes. One of those processes calls test1, the other does not. Then both processes fork again, creating a total of four processes; two of those processed call test2.

perl -E'sub Monkey::do{say$_,for@_,do{($monkey=[caller(0)]->[3])=~s{::}{ }and$monkey}}"Monkey say"->Monkey::do'

In reply to Re: parallel functions by tobyink
in thread parallel functions by micmac

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