Allison Randal wrote:
Presumably the handled opcode will remove the exception Task from the
scheduler and resume execution at the appropriate point. Presumably
also the declining to handle an exception (the replacement for
rethrow) will cause the scheduler to move to the next exception
Excellent, the patch from ticket 52214 works.
Walter
Jonathan Worthington schrieb:
Allison Randal wrote:
Will Coleda schrieb:
Can we get an idea of how many parrot hackers are planning on
attending YAPC::EU this year? (will be held in Copenhagen, Denmark, on
13-15 August 2008
Bernhard Schmalhofer wrote:
Is there enough critical mass that
# New Ticket Created by William Herrera
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/* compiler.h
* Copyright (C) 2007-2008, The Perl Foundation.
* SVN Info
*
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Hello.
There is implementation of kv method for rakudo.
Only
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The MS C compiler produces numerous warnings about not all paths return a
value
John M. Dlugosz wrote:
The synopses are contradictary over the way 'constant' works. First it says
that it is a declarator like 'my'.
That's what STD.pm says:
token scope_declarator:my { sym scoped {*} }
token scope_declarator:our { sym scoped {*} }
token scope_declarator:state
Moritz Lenz moritz-at-casella.verplant.org |Perl 6| wrote:
Then in S12 it shows
my constant ...
and
our constant ...
that is, independant from the my or our declarator.
I grep'ped STD.pm tentatively for other occurrences of 'constant', and
couldn't find where that should be
I want to review and collect the wisdom of what has been discussed before.
Someone mentioned this the other day, as being a significant consensus. But I
can't find anything in the forum archives.
Can someone point to the discussion, position papers, etc.?
--John
Perl 6 doesn't have references anymore, it has captures. So, what does
the following mean:
@x = foo bar;
$a = [1, 2, [EMAIL PROTECTED];
I imagine that the 3rd element of the Array is itself an Array, and is
the same object that is bound to @x. But captures are lazy
context-sensitive
John M. Dlugosz wrote:
Perl 6 doesn't have references anymore, it has captures. So, what
does the following mean:
@x = foo bar;
$a = [1, 2, [EMAIL PROTECTED];
I imagine that the 3rd element of the Array is itself an Array, and is
the same object that is bound to @x. But captures are lazy
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