Re: (tye)Re: Calling subroutine in a package from other perl sub.

Yes. This is a good explanation of the diferences between shift and accessing @_ directly. After I shift, the alias is gone from @_. So its obvious I cant modify the original value.

However, things dont exactly work like this though. And an alias is not the same as a reference. Correct? Because if test($x,$y,$z) does @_ = (\$x, \$y, \$z) then in then in the code you would have to dereference the variable after you shift them to access them. They would be references not scalar data. my $x = shift @_; print $$x;

zzSPECTREz

NOTE: I appreciate your responses. You allways get me thinking.

Comment on Re: (tye)Re: Calling subroutine in a package from other perl sub.

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(tye)Re2: Calling subroutine in a package from other perl sub. by tye (Sage) on Jan 17, 2001 at 22:19 UTC
Yes, exactly! After some thought, I need better (or just more) terminology. Let us say that you use the \ and -> operators with "Perl references". Then let us say that Perl itself makes heavy use of a type of reference internally but we'll call these internal references "pointers" since that is probably less confusing for this discussion. Now a "pointer" isn't what a C programmer would call a pointer since there is at least one more layer of abstraction that isn't important for this discussion (so a "pointer" contains a C pointer plus a few other things). So a Perl scalar variable is really a name for a "pointer" to a scalar value (called an SV in the source code). So $x points to some SV structure that contains $x's flags saying what type(s) of value(s) $x currently has and contains those values. And a Perl array is an array of "pointers" to SVs. So calling test(@v) copies the "pointers" from @v to @_ so we end up with `$v[0]` and `$_[0]` pointing to the same SV structure and changing a value via one also changes the value seen via the other. The two are said to be aliases of each other. But doing @_= @v causes new SV structs to be allocated for @_ and the values from @v's SVs are copied to @_'s new SVs. And so doing "print $x" involves an "automatic dereference" of $x's "pointer". Now a "Perl reference" is a(n) SV that contains a "pointer". You can watch aliases in work via: `sub refd { print join( " ", \(@_) ), $/; } my @v= qw( a b c ); print "@v: ", join( " ", \(@v) ), $/; refd( @v ); refd( $v[0] ); refd( @v[1,2,0] ); refd( "$v[1]" );` [download] You'll note that you get the same references except in the last case. - tye (but my friends call me "Tye")	[reply] [d/l] [select]

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(tye)Re2: Calling subroutine in a package from other perl sub.
by tye (Sage) on Jan 17, 2001 at 22:19 UTC

Yes, exactly!

After some thought, I need better (or just more) terminology.

Let us say that you use the \ and -> operators with "Perl references". Then let us say that Perl itself makes heavy use of a type of reference internally but we'll call these internal references "pointers" since that is probably less confusing for this discussion. Now a "pointer" isn't what a C programmer would call a pointer since there is at least one more layer of abstraction that isn't important for this discussion (so a "pointer" contains a C pointer plus a few other things).

So a Perl scalar variable is really a name for a "pointer" to a scalar value (called an SV in the source code). So $x points to some SV structure that contains $x's flags saying what type(s) of value(s) $x currently has and contains those values.

And a Perl array is an array of "pointers" to SVs. So calling test(@v) copies the "pointers" from @v to @_ so we end up with $v[0] and $_[0] pointing to the same SV structure and changing a value via one also changes the value seen via the other. The two are said to be aliases of each other.

But doing @_= @v causes new SV structs to be allocated for @_ and the values from @v's SVs are copied to @_'s new SVs.

And so doing "print $x" involves an "automatic dereference" of $x's "pointer".

Now a "Perl reference" is a(n) SV that contains a "pointer".

You can watch aliases in work via:

sub refd {
    print join( " ", \(@_) ), $/;
}
my @v= qw( a b c );
print "@v: ", join( " ", \(@v) ), $/;
refd( @v );
refd( $v[0] );
refd( @v[1,2,0] );
refd( "$v[1]" );
[download]

tye

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