On Wednesday 12 July 2006 11:27, Patrick R. Michaud wrote:
The perl6 compiler has a custom string type, currently called
Perl6Str. What's the canonically correct mechanism for creating
an object of that type?
$P0 = new 'Perl6Str'
$P0 = new .Perl6Str
$P0 = new [ 'Perl6Str' ]
On Wed, Jul 12, 2006 at 11:36:56AM -0700, chromatic wrote:
On Wednesday 12 July 2006 11:27, Patrick R. Michaud wrote:
The perl6 compiler has a custom string type, currently called
Perl6Str. What's the canonically correct mechanism for creating
an object of that type?
$P0 = new
On Wed, Jul 12, 2006 at 01:55:39PM -0500, Patrick R. Michaud wrote:
On Wed, Jul 12, 2006 at 11:36:56AM -0700, chromatic wrote:
On Wednesday 12 July 2006 11:27, Patrick R. Michaud wrote:
The perl6 compiler has a custom string type, currently called
Perl6Str. What's the canonically correct
On Wed, Jul 12, 2006 at 01:55:39PM -0500, Patrick R. Michaud wrote:
On Wed, Jul 12, 2006 at 11:36:56AM -0700, chromatic wrote:
On Wednesday 12 July 2006 11:27, Patrick R. Michaud wrote:
$P0 = new 'Perl6Str'
I tend to use:
.local int str_type
str_type = find_type [
On Wed, Jul 12, 2006 at 12:15:07PM -0700, Chip Salzenberg wrote:
- If another HLL wants to create a Perl6Str, how does it do it?
- If another HLL wants to create a subclass of Perl6Str...?
I just realized that I misinterpreted these questions. I thought that the
first question was
