What error do you get?

When dealing with syntax errors, (and also almost every other error), I find it most helpfull to eliminate things until the error goes away.

Have you looked at the examples in the split documentation and compared them to your code?

Something like

knoppix@Microknoppix:~$ perl -E '
> $str = q{abc1)def2)g(h)i3)jkl};
> say for split m{(?<=\d)\)}, $str;'
abc1
def2
g(h)i3
jkl
knoppix@Microknoppix:~$
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perhaps?

$hh="1)foo2)bar3)blabla";
@nn=split /\d\)/, $hh;
print join ':', @nn;
print "\n";
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@li=split (\d\), $l);#  - escaping the )

You need to enclose your regex in slashes /.../

It is possible to delimit regular expressions with other chars, and I though that perhaps the OP was stumbling over this feature of perl, and accidentally using round brackets as a delimiter, so I did a quick test

split ( ), '48 65 6c 6c 6f 2c 20 77 6f 72 6c 64 21'
split m( ), '48 65 6c 6c 6f 2c 20 77 6f 72 6c 64 21'
split qr( ), '48 65 6c 6c 6f 2c 20 77 6f 72 6c 64 21'
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The second two expressions split the list, but without anything to tell perl that the first is a regular expressions it was not treated as one. I also tried:

split (\s+), '48 65 6c 6c 6f 2c 20 77 6f 72 6c 64 21'

But that resulted in a syntax error unrelated to split.

FWIW split ( ), 'a b c' is parsed as split(), 'a b c '. So it's a call to split with no arguments. That call is part of a list, and the string literal is the last part of the list. Certainly not what the OP wants.

Perl 6 - second systems done right

To "split the string at the character that precedes \d\)", use a zero-width lookahead for the digit and paren:

split /.(?=\d\))/, $1
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