$_ is .. what of the caller?

$_ is $_. No telling how the caller is using it.

And am I to understand that in sub a, messing with @_, regardless if the caller passed any references as arguments, is potentially messing with the caller's arguments?

Easy to test...

Not quite. Messing with the elements of @_ will. Adding elements and removing elements (including assigning to @_ specifically) won't. 

foo($x, $y+1, $z);

[download]
is basically equivalent to 
{
    local @_;
    alias $_[0] = $x;
    alias $_[1] = $y+1;
    alias $_[2] = $z;
    &foo;
}

[download]

Changing $_[0] in foo will change $x, for example.

In reply to Re^5: using shift and $_ in loop by ikegami
in thread using shift and $_ in loop by leocharre

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