Assuming the following are true:

    A: "if" is now a normal function

    B: "foo() + 3" is (foo) + 3, foo doesn't get 3.

Then does that mean we're stuck with:

    C: "if($foo) { say 'foo' }" being a syntax error?

I currently think A is wrong. Am I right?

Juerd
--
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There will always be various control constructs that cannot be written
as an equivalent function. Otherwise, there is no way to write the
higher level ones.

I suspect that "goto" and "if" are on this list. Possibly they are the
only things on this list, but "return" is a strong contender.

Additionally, any predicate form cannot be a "normal function". You
might be able to define an &infix:<for>, but getting to where you don't
need the {}'s on the LHS block, and getting a proper list on the RHS
gets rather abnormal.

Something that crossed my mind while writing this: Does

    for { say } <== @a;

Work?

-- Rod Adams

Rod Adams skribis 2005-04-25 14:37 (-0500):

Not a problem, because we're using something to bootstrap, and the perl
6 you'll be using to run your Perl 6 scripts will be in compiled form,
free of any need to parse itself.

Self-definition is a wonderful thing.

If for's just a function that takes a slurpy list, then I see no reason
why not.

Juerd
--
http://convolution.nl/maak_juerd_blij.html
http://convolution.nl/make_juerd_happy.html
http://convolution.nl/gajigu_juerd_n.html

Almost.  It's a statement-level form, which looks like a normal function
except for the statement: prepended on its name.  Such constructs (which
include for, while, the whole gang) have a few special properties:

* They can't be used intra-expression.

    say 3 + if foo() { 4 } else { 5 }    # error!

* An opening brace anywhere (not inside brackets) in operator position
  gets passed to them:

    sub foo (*@_);
    sub bar (*@_);

    if foo 1, bar 2 { ... }
                    ^ belongs to the if

You could use if() intra-expression like so:

    say 3 + &statement<if else>().({foo()}):{4}:{5};

(Ick).

Luke

: Something that crossed my mind while writing this: Does
:
:    for { say } <== @a;
:
: Work?

Nope.  The brackets where a term is expected would be misconstrued
as an argument to the "for".  Maybe we need an @= for a placeholder:

    for @= { say } <== @;

Or maybe we could just make () serve that purpose:

    for () { say } <== @;

Larry