%a ^:union[op] %b
%a :foo[op]:bar %b
I think that any operators over 10 characters should
be banished, and replaced with functions.
I'd agree with that. In fact probably anything over 4,
and even 4 is seriously pushing it.
I'll clarify that I am talking here about using
On 31 Oct 2002 at 0:40, John Williams wrote:
On Wed, 30 Oct 2002, Me wrote:
%a ^:union[op] %b
%a :foo[op]:bar %b
I think that any operators over 10 characters should be banished, and
replaced with functions.
I don't think there should be any upper limit for operator-lengths.
Uri Guttman writes:
%hash1.values [+]= %hash2{%hash1.keys} ;
but here is exactly example analogous to
my Dog $x = new Dog .
which was discusse dand turned to
my Dog $x .= new ;
It's (almost) clear what you want when you write
%hash1 [+]= %hash2 ;
so why to screen the
Me writes:
%a ^:union[op] %b
%a :foo[op]:bar %b
I think that any operators over 10 characters should
be banished, and replaced with functions.
I agree. But I think that we can get away here with just hash
properties , just like hash behaviour in regexps is
On Thu, 31 Oct 2002, Markus Laire wrote:
: I don't think there should be any upper limit for operator-lengths.
There will never be any official limits. Perl is not about arbitrary
limits. But I will tell you that I only added = to Perl 5 because
I knew there would never be a == operator. We'll
Austin Hastings writes:
--- [EMAIL PROTECTED] wrote:
I agree. But I think that we can get away here with just hash
properties , just like hash behaviour in regexps is controlled by
properties .
e.g.
union:
(%a,%b) ^is no_strict_keys ;
(%a %b) ^is default_value
Austin Hastings writes:
but I am not shure ...
sure
thanks . sorry that I write so badly . I'll try to be better .
(Unless you do this on purpose :-)
Cheers,
=Austin
__
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Me writes:
union:
intersection :
How would this work for hashes with differing properties?
%a ^is strict_keys;
%b ^is no_strict_keys;
What would happen?
in the resulting hash only ( and all ) keys of %a will be present.
because %b *admits* unknown keys but %a
Larry Wall writes:
On Thu, 31 Oct 2002, Me wrote:
: That's one reason why I suggested control of this sort
: of thing should be a property of the operation, not of
: the operands.
I think that by and large, the operator knows whether it wants to
do union or intersection. When
On Thu, 31 Oct 2002, Me wrote:
: That's one reason why I suggested control of this sort
: of thing should be a property of the operation, not of
: the operands.
I think that by and large, the operator knows whether it wants to
do union or intersection. When you're doing +, it's obviously
Me writes:
union:
intersection :
%a ^is strict_keys;
%b ^is no_strict_keys;
in the resulting hash only ( and all ) keys of %a will be present.
because %b *admits* unknown keys but %a does not.
Yes, but the general case is that one wants to be
--- Larry Wall [EMAIL PROTECTED] wrote:
On Thu, 31 Oct 2002, Markus Laire wrote:
The really great thing about the French quotes is that they visually
keep the user aware of the composition. «+=» is obviously a variety
of
+=, whereas ^+= is not obvious, though shorter. (Square brackets are
Proposal : Vector operations for Hashes
%a ^{+} %b
the direct generalization would be as follows :
a ^[op] b
for a ; b -
$x ; $y {
$x op $y
}
%a op %b
# find common keys
k = keys hash{ map { { _=1 } } *( %a.keys , %b.keys ) };
#return hash
On Wed, 30 Oct 2002 [EMAIL PROTECTED] wrote:
: # find common keys
: @k = keys hash{ map { { _=1 } } *( %a.keys , %b.keys ) };
It seems a bit odd to privilege the intersection over the union.
Larry
!! Sorry, in previos post I had a mistake
Proposal : Vector operations for Hashes
(
this is just a joke
%a ^{+} %b
)
the direct generalization would be as follows :
a ^[op] b
for a ; b -
$x ; $y {
$x op $y
}
%a ^[op] %b
# find
So is it usefull to make sence out of these :
hash ^[op] hash
hash ^[op] array
hash ^[op] scalar
array ^[op] array
array ^[op] scalar
Actually ,
I just want to understand , why so much accent is put on vectorizing
atomic operations with arrays . It seems that hashes are at least as
hash ^[op] hash
...
array ^[op] scalar
ie, generally:
term ^[op] term
what to do if a, b in a ^[op] b have different length
what to do if %a, %b in %a ^[op] %b have not the same set of keys
what to do in %a ^[op] a
[what to do] resolved by hash property :
I'd expect adverbs
fc == [EMAIL PROTECTED] writes:
fc So is it usefull to make sence out of these :
fc hash ^[op] hash
fc hash ^[op] array
fc hash ^[op] scalar
fc array ^[op] array
fc array ^[op] scalar
well, you can't mung the keys to a hash, just the values. so why not
just use the values
On Wed, 30 Oct 2002, Me wrote:
%a ^:union[op] %b
%a :foo[op]:bar %b
I think that any operators over 10 characters should be banished, and
replaced with functions.
~ John Williams
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