* shmem <[EMAIL PROTECTED]> [2007-12-10 12:55]:
> Well, if we were to return an array reference instead of a hash
> ref,
>
> sub [EMAIL PROTECTED],shift}
>
> works. Why does the shift get executed before an array
> reference is constructed - but not if a hashref is constructed
> - from an array?
It probably doesn’t. Either way, evaluation order is a red
herring: your construction will work regardless of which
subexpression is evaluated first!
You managed to confuse yourself. Congratulations. :-) (Or you
are underhandedly trying to confuse us. In that case, sorry bub,
better luck next time. :-) )
What happens is that `\` takes a *reference* to the array. And
of course when you do that, any modifications of the referent,
including *subsequent* modifications, will be seen by anyone who
holds a reference to it.
In contrast, the hash constructors make a *copy* of the array,
and that copy will not be affected but subsequent modifications
to the source array.
Anyone who has trouble following should consider the difference
between the following two:
sub new { bless [EMAIL PROTECTED], shift }
# vs
sub new { bless [EMAIL PROTECTED], shift }
Consider how many arrays are involved in either case, and which
one is affected by the `shift`.
--
*AUTOLOAD=*_;sub _{s/(.*)::(.*)/print$2,(",$\/"," ")[defined wantarray]/e;$1}
&Just->another->Perl->hack;
#Aristotle Pagaltzis // <http://plasmasturm.org/>
