On Fri, 4 Nov 2005, TSa wrote:
"derived" from C. But Perl's C<++> already allows an extended syntax wrt
that of those other languages, that is: I'm not really sure, but I don't
[snip]
Actually I'm not sure if Perl 6 allows
($u *= 5)++;
Since I wrote "already", I was referring to Perl5...
because the return value of *= is the rvalue 5 and *not* $u as
an lvalue. Correct me if I'm wrong. Otherwise how would the
expanded form
...but I don't have the slightest idea whether Perl6 will allow it or not.
OTOH I feel like asking _why_ should it not? I mean: why "should" the
return value of *= be the rvalue 5 and not $u as an lvalue, which would
still return its new rvalue in rvalue context and be assignable in lvalue
context?
behave when extented by yet another assignment
($y = $u = $u * 5)++; # $y++ or $u++
I think $y++ since it has been assigned the return value of the expression
$u = ... which happens to be 5.
and recollapsed as
($y = $u *= 5)++;
say $y; # 6?
it is definitely so in Perl5:
$ perl -le '$u=1; ($y=$u*=5)++; print $y'
6
At that stage you could drop the application to x and store the resulting
action in another variable C := (A * B) for later application. Writing the
above with Perl6 sigils would come out as
&A * &B * $x;
and then
&C = &A * &B;
Would not this be a sort of currying, thinking of this as a ternary
operator that takes &A, &B and $x?
is a calculated code object just like
$y = $a * $b;
is a calculated value. This is what I would call first class operators!
But it depends on what C<*> is here. If that instead of being a
multiplication of matrices which is (a somewhat specialized composition
law but essentially boils down to a) composition of function, were a
generic composition of functions, then yes: I'd like an easy enough form
of syntactic sugar for that.
I remember having asked something akin to this in the context of what I
would have considered a very natural syntax for doing something (but
$Larry seemed to disagree) which implied map()ping a list of scalars into
a list of closures to 'chain' (reduce?) them with a composition operator.
Briefly: ouch, This is what I would call first class operators too!
Michele
--
Whoa! That is too weird! I asked around among the math
faculty here and it turns out that _every one's_ wife
is married to a mathematician!
- Dave Rusin in sci.math, "Re: Genetics and Math-Ability"
