David Whipp wrote:
$b = 7, 6, 5
b = 7, 6, 5
I understand that C's *interpretation* of the comma operator will be expunged from
Perl 6. But unless comma's *precedence* is also changing, neither of those statements
would build a list with three elements.
It seems to me that
$b = 7, 6,
It seems that the fundamental problem is the dichotomy between
a scalar, and a list of 1 elem. Thus, we want
$a = 7
to DWIM, whether I mean a list, or a scalar. Seems to me that
the best way to solve a dichotomy is to declare it to not to
be one: a scalar *IS* a list of one element. The only
On Tue, Sep 24, 2002 at 11:47:16AM -0700, David Whipp wrote:
It seems that the fundamental problem is the dichotomy between
a scalar, and a list of 1 elem. Thus, we want
$a = 7
to DWIM, whether I mean a list, or a scalar. Seems to me that
the best way to solve a dichotomy is to declare
On Tue, 2002-09-24 at 14:47, David Whipp wrote:
It seems that the fundamental problem is the dichotomy between
a scalar, and a list of 1 elem. Thus, we want
After the first couple of messages, that was really no longer *my*
concern, but I can't speak for others. My concern was mostly that
2. Scalar assignment.
my $a;# 1.
$a = X;
my $a;# 3.
($a) = X;
These should all do the same thing, regardless of X.
Consider:
$a = (1);
and
($a) = (1);
5. Assignment to arrays and lists.
$a = (1, 2, 3); # Same as Perl 5's $a = [1,2,3];
$a = (1)
In a message dated Tue, 24 Sep 2002, Mike Lambert writes:
Consider:
$a = (1);
and
($a) = (1);
Yes? They both do the same thing--set $a to 1. It looks like the bottom
one is a list assigned to a list, but that might be optimized out, as it
doesn't matter.
5. Assignment to arrays and
From: Jonathan Scott Duff
$b = 7, 6, 5
b = 7, 6, 5
Again, both create identical objects, under different
interfaces. But now we have a problem with +$b: what should
this mean? To be consistant with +$a (above), I would
suggest that it simply returns the sum of its elements
On Tue, 24 Sep 2002, Mike Lambert wrote:
$a = (1, 2, 3); # Same as Perl 5's $a = [1,2,3];
$a = (1) should then do $a = [1], according to the above.
This implies that:
($a) = (1) implies that $a is [1], something I don't particularly agree
with.
You may be missing the change in the
[EMAIL PROTECTED] (Trey Harris) writes:
May I suggest that we start with some DWIMmy examples
Sam sat on the ground and put his head in his hands. 'I wish I had
never come here, and I don't want to see no more magic,' he said, and
fell silent.
--
I hooked up my accelerator pedal in my car to
On Mon, 23 Sep 2002, Jonathan Scott Duff wrote:
On Mon, Sep 23, 2002 at 04:58:55PM -0400, Trey Harris wrote:
for (1,(a,b,c),3 { ... }
and
for 1,(a,b,c),3 { ... }
Now that I've ventured away from DWIMs and more into WIHDTEMs (What In
Hell Does This Expression Mean), is the
In a message dated Mon, 23 Sep 2002, Luke Palmer writes:
Y'all have it backwards.
[1,*[2,[3,4,5]],6] # [1,2,[3,4,5],6]
[1,*[2,*[3,4,5]],6] # [1,2,3,4,5,6]
Flat flattens outwards, not inwards.
Ah. *slaps head* of course. That makes much more sense.
On Mon, 2002-09-23 at 16:58, Trey Harris wrote:
4. Numeric value.
The progression spoken about at great length previously:
+()# == 0
+(0) # == WHAT? 0? 1?
+(0,1) # == 2
+(0,1,2) # == 3
+(0,1,2,3) # == 4
+(0,...,n) # == n + 1
is largely
Replying to myself to clear a few things up...
In a message dated Mon, 23 Sep 2002, Trey Harris writes:
2. Scalar assignment.
my $a;# 1.
$a = X;
my $a;# 2.
$a = X;
my $a;# 3.
($a) = X;
my($a) = X; # 4.
my($a) = (X); # 5.
On Mon, 23 Sep 2002, Trey Harris wrote:
So then, I think if there's just some clarification about how one-tuples
are formed, I think everything I wrote in my earlier mail can DWIM
correctly. There seems to be no magic here, quotations from LoTR to the
contrary. :-)
Your post was very
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