On Wed, Jun 06, 2001 at 01:37:23AM -0500, Me wrote:
BD languages
What's BD?
Bondage and Discipline, scum! You're not a good enough programmer to
be trusted not to make mistakes! Now drop and give me fifty!
--
Michael G. Schwern [EMAIL PROTECTED]http://www.pobox.com/~schwern/
On Wed, Jun 06, 2001 at 01:37:23AM -0500, Me wrote:
Larry's MMV on that ;-)
Man I really need to get up to speed with these
acronyms. I know YMMV, is MMV a distant
cousin perhaps?
Same idea, except it's Larry's Milage in question, rather than Yours.
dha
--
David H. Adler - [EMAIL
And, if this is so, then isn't it impossible to have useful
stricture about variable properties, because any given
reference to a property might be instead a value property
unknown to the compiler?
Yes.
So:
You can't have (variable or value) property stricture.
Do
Me wrote:
I.Found your notion of a sealed off namespace
intriguing. I have no idea what it meant just yet;
I'm going to go read and think about it now.
I'll pitch some syntax:
# prevent modification to %reflexive:: like so:
package reflexive is closed;
# allow it
Me wrote:
Question 1:
Afaict, even with use strict at its most strict, perl 6
can't (in practice) complain, at compile time, if
$foo.Foun
refers to an undeclared Foun.
Right?
it is already detectable. from perldoc perlref:
Perl will raise an exception if you
Afaict, even with use strict at its most strict, perl 6
can't (in practice) complain, at compile time, if
$foo.Foun
refers to an undeclared Foun.
it is already detectable. from perldoc perlref:
Perhaps for perl 5, but, aiui, Damian confirmed
that my thinking about
I apologize. I royally screwed up my original post.
I had meant to ask two minor specific yes/no answer
type questions about properties and stricture, that
were mutually unrelated. Instead I asked one major
open ended one.
In the hope that I haven't completely blown any
chance of getting
On Tue, Jun 05, 2001 at 04:38:24PM -0500, Me wrote:
Question 1:
Afaict, even with use strict at its most strict, perl 6
can't (in practice) complain, at compile time, if
$foo.Foun
refers to an undeclared Foun.
Right?
Can't you hear the low roar from the strong-typing
Question 1:
Afaict, even with use strict at its most strict, perl 6
can't (in practice) complain, at compile time, if
$foo.Foun
refers to an undeclared Foun.
It could certainly warn you, but it can't object fatally since there's
always the
Question 2:
Afaict, even with use strict at its most strict, perl 6
can't (in practice) complain, at compile time, if
$foo.Foun
refers to an undeclared Foun.
It could certainly warn you
Consider the code:
my $foo = 1 is Found;
On Tue, Jun 05, 2001 at 04:38:24PM -0500, Me wrote:
Question 1:
Afaict, even with use strict at its most strict, perl 6
can't (in practice) complain, at compile time, if
$foo.Foun
refers to an undeclared Foun.
Right?
Can't you hear the low roar from the
Consider the code:
my $foo = 1 is Found;
bar($foo);
sub bar { my $baz = shift; if ($baz.Found) { ...} }
Does the value of $baz have the Found property?
Yes.
If so, does the compiler know that?
No. Because it only has the property at
Consider the code:
my $foo = 1 is Found;
bar($foo);
sub bar { my $baz = shift; if ($baz.Found) { ...} }
Does the value of $baz have the Found property?
Yes.
If so, does the compiler know that?
No. Because it only has the property
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