I'm afraid you can't do this in your Perl program.

It's about context... and happens in the depths of Perl's toke.c. After the name of a subroutine, '<' might be the start of the sub's parameter list, either some diamond or a heredoc. Unfortunately, the code handling these does not have a fallback to "ok, let's try an operator instead" if it's neither. A '>', on the other hand, does not have any ability to produce a parameter list, and is only tested as an operator.

All the workarounds seen here - bla(), (bla), &bla, sub bla() - move Perl into a context where only operators are permitted, and I can't come up with anything else.

BTW: Overloading doesn't play any role here. It only affects code which has been successfully parsed.

Thanks for checking the parser's code! haj++

> BTW: Overloading doesn't play any role here. It only affects code which has been successfully parsed.

Yeah I'm aware of this, but wanted to give some context (no pun intended ;-)

Cheers Rolf
_{(addicted to the Perl Programming Language and ☆☆☆☆ :)
Wikisyntax for the Monastery}

Hi LanX,

Try it like this:

  DB<13> &bla < bla
works at ...
[download]

I am not 100% clear on why this works, I would appreciate it if someone else might be able to explain it to us :D

Best,

Jim

Indeed thanks! :)

see perlsub for the explanation:

&NAME;       # Makes current @_ visible to called subroutine.

It's a syntax to pass parameters through, so it can't accept new arguments.

  DB<27> sub tst { [@_] }

  DB<28> sub outer { &tst }


  DB<29> x outer(1,2,3)
0  ARRAY(0x2f700e8)
   0  1
   1  2
   2  3

  DB<30> x &tst 1,2,3
Number found where operator expected at
[download]

Cheers Rolf
_{(addicted to the Perl Programming Language and ☆☆☆☆ :)
Wikisyntax for the Monastery}

update

for clarification, while interesting, this doesn't answer my question and side effects are too weird...

(bla) < bla
[download]

($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,
[download]

Thanks! :)

That's what I meant with "or embrace it in brackets" and it's my preferred workaround at the moment...

But I'd prefer to disable <> locally...

Cheers Rolf
_{(addicted to the Perl Programming Language and ☆☆☆☆ :)
Wikisyntax for the Monastery}

IO::All comes to mind as a module which implemented overloading of "<" and whose technique might work for you. Didn't dig into that though, sorry for laziness.

update: couldnt resist... scratch that:

perl -MIO::All -E 'sub bo {"blorf"} sub ba {"bar"} ba < bo'
Unterminated <> operator at -e line 1.
[download]

Nope.

To make that work - I fear (just guessing!) - you'll have to go all the way back to perly.c and modify that, fighting with the magic of "<>".

update2: this is bug smell to me.

update3: it looks like in the expression

foo < bar
[download]

the foo thingy must resolve to something that doesn't consume what follows it on the right, overloading in effect or not, to be parsed as you want it to be parsed:

$ perl -E 'say < 2'
Unterminated <> operator at -e line 1.
$ perl -E 'say() < 2'

$ perl -E '(say) < 2'

$ perl -E 'say scalar(say) < 2'

1
$ perl -E 'sub foo(){1} say foo < 2'
1
[download]

So, no. Using the LHS of < as a function which does or does not take arguments isn't possible, I guess.