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Postings from October 2015
On Mon, Oct 12, 2015 at 12:54:31PM -0400, Eric Brine wrote:
> On Mon, Oct 12, 2015 at 6:22 AM, Dave Mitchell <davem@iabyn.com> wrote:
>
> > On Wed, Oct 07, 2015 at 03:57:26PM -0400, Eric Brine wrote:
> > > On Wed, Oct 7, 2015 at 4:24 AM, Dave Mitchell <davem@iabyn.com> wrote:
> > >
> > > > 2. Make when's behaviour entirely dependent on the *compile-time* class
> > > > of of its arg. In particular, make 'when (FOO)' have exactly the
> > following
> > > > behaviours:
> > > >
> > > > FOO condition evaluates to
> > > > -------------- ----------------------
> > > > literal undef !defined($_)
> > > > numeric const $_ == FOO
> > > > literal pattern $_ ~= FOO
> > > > all else $_ eq FOO
> > > >
> > > > 3. when { FOO } has condition FOO.
> > > >
> > > > For example:
> > > >
> > > > use constant ME => "davem";
> > > > when (ANS) # equivalent to when { $_ == 42 }
> > > >
> > >
> > > According to your table, shouldn't that be "when { $_ eq ANS }"? ANS
> > could
> > > have any value, so when (ANS)" is no different than "when ($x)". There's
> > no
> > > "compile-time class" available to study.
> >
> > No, its specifically if FOO is a compile-time constant. At compile time
> > there is no difference between
> >
> > when(42)
> >
> > and
> >
> > use constant FOO;
> > when(FOO);
> >
> > They have both been compiled to a single OP_CONST with a constant SV
> > attached that can be examined at compile-time for numness.
>
>
> It doesn't matter what code it *results* in.
>
> use constant FOO => $x;
> when (FOO)
>
> is no different than
>
> when ($x)
>
> and we all agree the latter is not workable. FOO is the still the result of
> an arbitrary expression, even if the expression is evaluated at
> compile-time. This differs from
>
> when ("x")
> when (4)
>
> because that bases behaviour on the *source* code.
This is Perl. You can't even trust the source code:
use constant;
use overload::constant integer => sub {"$_[0]"};
when (3) {...} # What does this do? 'eq' or '=='?
Abigail
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