HaloO,
Larry Wall wrote:
Mostly that means that rw will cause autovivification and ref won't.
As noted in the 'assignable mutators (S06/Lvalue subroutines)' thread
I assume that scalar containers as such cannot be autovivified, right?
But not yet existing entries in an array or hash can be
In S06, what is the difference between is ref and is rw? The text says
that the rw may be converted to an lvalue, and that ref must already be. But
what is that supposed to mean?
--John
On 2008 May 10, at 21:46, John M. Dlugosz wrote:
In S06, what is the difference between is ref and is rw? The
text says that the rw may be converted to an lvalue, and that ref
must already be. But what is that supposed to mean?
At a guess, is rw makes a parameter variable into a local
On Sun, May 11, 2008 at 01:46:57AM -, John M. Dlugosz wrote:
: In S06, what is the difference between is ref and is rw? The text says
that the rw may be converted to an lvalue, and that ref must already be. But
what is that supposed to mean?
Mostly that means that rw will cause
On Sat, May 10, 2008 at 09:51:26PM -0400, Brandon S. Allbery KF8NH wrote:
On 2008 May 10, at 21:46, John M. Dlugosz wrote:
In S06, what is the difference between is ref and is rw? The text
says that the rw may be converted to an lvalue, and that ref must already
be. But what is that
Brandon S. Allbery KF8NH allbery-at-ece.cmu.edu |Perl 6| wrote:
On 2008 May 10, at 21:46, John M. Dlugosz wrote:
In S06, what is the difference between is ref and is rw? The
text says that the rw may be converted to an lvalue, and that ref
must already be. But what is that supposed to
