"@_ is a reference, and change its value will be seen in the calling func"

Er - NO. From perldoc perlsub

The array @_ is a local array, but its elements are aliases for the actual scalar parameters.

So, you are wrong on 2 counts:

• @_ is an array, not a reference
• Since it is local, changing @_ will NOT change the caller

I wanted to understand what was happening, what affected the caller and what didn't so I put a short script together. I may have missed ways to get at @_ and I would be interested if any omissions were pointed out.
Read more... (1353 Bytes)

The output produced is

At initialisation
     Contents are 1 2
In assignTo - assign to @_
     Contents now 1 2
In listArgs - list of lexicals
     Contents now 1 2
In loopOver - loop over @_
     Contents now 2 3
In refTo - take reference to @_ elements
     Contents now 3 4
In shiftArgs - lexicals via shift
     Contents now 3 4
In useDirect - increment elements directly
     Contents now 4 5
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A very interesting topic that has cleared up some misunderstandings.

Cheers,

JohnGG

Looks good (++).

Update:Ignore the rest of this node - as johngg points out, below - this code is equivalent to his assignTo sub.

If you like, you could add what "Anon Monk" implied was possible:

sub Modify_param_array{
   print qq{In Modify_param_array - change @_\n};
   @_ = qw[New contents. The caller never sees this];
}
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