Carl Jolley wrote:
> On Fri, 8 Feb 2002, $Bill Luebkert wrote:
>
>
>>Carl Jolley wrote:
>>
>>
>>>I you want a subroutine to return a reference then the only way
>>>for that to happen is that the subroutine has to return the
>>>reference. There is no way that I know of, external to a subroutine,
>>>to get reference to the subroutine's return value. You of course
>>>can get the subroutine's return value and create a reference to
>>>it but that's not exactly what I think you are looking for.
>>>
>>I believe you can do what he is asking, but you must know what the
>>routine is returning to do it properly.
>>
>>For scalar ref:
>> $self->{_result} = \$some_obj->some_method;
>>
>>vs array ref:
>>
>> $self->{_result} = [$some_obj->some_method];
>>
>>Of sourse technically you are creating a ref to the return value.
>>
>>
>
> I'm not sure what
>
> $self->{_result} = \$some_obj->some_method;
>
> generates (? a reference to the code?) but it seems to me that:
>
> $self->{_result} = [$some_obj->some_method];
>
> would have _result be reference to a single element anonymous array
> containing whatever scalar value that the method returned. I don't belive
> that technically (or otherwise) it would generate a reference to the value
> returned by the method even if the method returned an array reference.
Did you try it ? Didn't use an object on left side since it isn't necessary
to prove the point.
use strict;
use Data::Dumper; $Data::Dumper::Indent=1;
my $obj = new fubar;
my $foo = \$obj->foo;
print "\$\$foo = $$foo\n";
print &Data::Dumper::Dumper($foo);
print "\n";
my $bar = [$obj->bar];
print "\@\$bar = @$bar\n";
print &Data::Dumper::Dumper($bar);
package fubar;
sub new {
my $class = ref ($_[0]) ? ref (shift) : shift;
my $self = { };
return bless $self, $class;
}
sub foo {
return 4;
}
sub bar {
return (1, 2, 3);
}
__END__
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