David Nicol wrote:
> On 12/10/05, $Bill Luebkert <[EMAIL PROTECTED]> wrote:
>
>>The fact that you can modify the vrbl passed to a sub from in the sub without
>>using a reference still doesn't mean that it's passed by reference.
>
> No. You are mistaken. That is exactly what passing by reference means.
In Perl a reference is created using this syntax (there are other alternate
ways to create a reference, but that's the normal way) : \$var
If you didn't use that syntax, it's not a reference. So we have a syntactical
disagreement. Just because you can use a vrbl like a reference doesn't make it
one (in the Perl sense). We're talking Perl references here.
> Example:
> sub Increment($){ $_[0]++ };
>
> If you want to protect your passed arguments from possible modification,
> make a copy of them before submitting them:
>
> my @args=($SomeVariable);
> &Increment(@args); # $SomeVariable will not be touched
You're still confusing the fact that an target argument can be modified
with the fact that a vrbl is a reference or not. Because you can modify
a vrbl indirectly through an argument name doesn't make it a true Perl
reference. This is just some special aliasing logic that allows this to
work in a sub. What's passed is the value rather than a reference to the
value. All you have to do to prove that is print it:
my $one = 1;
mysub ($one); # prints '1'
mysub (\$one); # prints 'SCALAR(0x23bd88)'
sub mysub { print "@_\n"; }
You have to speak in Perl terms rather than what you think of in non-Perl
terms - a reference is a very specific animal in Perl.
> You could even write a byvalue wrapper if you want, something like
>
> sub ByValue(\&;){
> my $coderef = shift;
> my @argvals = @_; # copying arg refs to new array, by value, or
> at least COW
> &$coderef(@argvals);
> };
That's not necessary - just shift the args into local vrbls and
you have the same thing - why complicate it. Quit thinking in terms
of basic and think in terms of Perl.
my $var = 1;
whatever ($one);
sub whatever {
my $local_var = shift;
or
my $local_var = $_[0];
<do stuff to $local_var and it doesn't modify the orig $var>
}
You can specifically test for a reference in a sub and handle both cases
as we showed earlier I believe.
my $one = 1;
mysub ($one);
mysub (\$one);
sub mysub {
my $var = shift;
if (ref $var) {
print "$$var\n";
} else {
print "$var\n";
}
}
Now both calls will print '1'.
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