Perl - lookahead assertion - perl

Question

I'm trying to describe a perl script for this purpose:
a = ~b & ~c; ==> a = (~b) & (~c);
a = ~b & (~c); ==> a = (~b) & (~c);
So I used lookahead assertions to insert parenthesis like this. Here is the test code.
#!/usr/local/bin/perl5 -w
use strict;
use warnings;
my $line;
my #lines;
#lines = (
"assign a = ~b & ~c;",
"assign a = (~b) & (~c);",
"assign a = ( ~b & ~c );",
"assign a = (b & ~c );"
);
foreach $line (#lines) {
print " $line\n";
$line =~ s/(?!\(\s*)~\w+(?!\s*\))/\($&\)/g;
print ">> $line\n\n";
}
It looks working with above examples. However, it's not working with this.
assign a = ~b & ~c;
>> assign a = (~b) & (~c); <== OK
assign a = (~b) & (~c);
>> assign a = (~b) & (~c); <== OK
assign a = ( ~b & ~c);
>> assign a = ( (~b) & ~c); <== X. I want ( (~b) & (~c));
assign a = ( ~b & ~c );
>> assign a = ( (~b) & ~c ); <== X. I want ( (~b) & (~c) );
Would you let me know how to fix the script? Thank you.

Answer 1

Your goal of using lookahead and lookbehind assertions doesn't really get you anything. Breaking up the code into two steps makes it easier in my opinion. One step to capture variables prefixed by ~, and the second part to see if they're surrounded by balanced parenthesis.
use strict;
use warnings;
while (<DATA>) {
chomp(my $src = <DATA>);
chomp(my $test = <DATA>);
$src =~ s{([(]?~\w+[)]?)}{
my $str = $1;
$str =~ /^\(.*\)$/ ? $str : "($str)";
}eg;
print "test $test\n";
print $src eq $test ? ' ok ' : ' FAIL! ';
print "$src\n";
}
__DATA__
Test:
a = ~b & ~c;
a = (~b) & (~c);
Test:
a = (~b) & (~c);
a = (~b) & (~c);
Test:
a = ( ~b & ~c);
a = ( (~b) & (~c));
Test:
a = ( ~b & ~c );
a = ( (~b) & (~c) );
results:
test a = (~b) & (~c);
ok a = (~b) & (~c);
test a = (~b) & (~c);
ok a = (~b) & (~c);
test a = ( (~b) & (~c));
ok a = ( (~b) & (~c));
test a = ( (~b) & (~c) );
ok a = ( (~b) & (~c) );

Answer 2

You can't easily do what you're asking using a single regular expression.
The problem is that there is no way to count the number of nested parentheses without writing a recursive regex pattern, so at the end of ~c simple regex cannot know whether how many parentheses are needed to close the expression.
This is possible with a more complex regex, but it would also be much easier to tokenize the string in a Perl loop.
Do you have to deal with stuff like a & ~b & c | (d | ~e & f)?

Answer 3

You can do this with one regex, here it is;
$line =~ s/(?|([^\(])(~\w+)(.)|(.)(~\w+)([^\)]))/$1\($2\)$3/g;
Your reqex wasn't doing what you thought.
$line =~ s/(?!\(\s*)~\w+(?!\s*\))/\($&\)/g;
the first part "(?!(\s*)~" will never match. Remember lookaheads and lookbehinds are zero width assertions. I like to think of them as matching the space in between the letters. (?!(\s*)~ means, that you want to match a "~" character, but in the space right before the "~"character, you want to lookahead and make sure you dont see a "(" and spaces. Well, if you are in the space right before a "~", you"ll never see a "(". And if your at a "(", the negative look ahead might fail to match (like you want) but you'd have never matched the "~" anyway.
You are trying to match if the character before is not a "(" AND the char after is not a ")". But what you want is to match if the character before is not a "(" OR the char after is not a ")". So you need a conditional branch, one to match if there is no "(" in front, and one to match if there is no ")" behind.
I used a condition branch, the (?| tells the engine to store the captured submatches like this;
(?|([^\\(])(~\w+)(.)|(.)(~\w+)([^\\)]))
$1 $2 $3 |$1 $2 $3
instead of this
([^\\(])(~\w+)(.)|(.)(~\w+)([^\\)]))
$1 $2 $3 |$4 $5 $6
I used (.) to make the ~\w part always $2, then just put a "(" ")" around the $2 in the output
my output
assign a = ~b & ~c;
assign a = (~b) & (~c);
assign a = (~b) & (~c);
assign a = (~b) & (~c);
assign a = ( ~b & ~c );
assign a = ( (~b) & (~c) );
assign a = (~b & ~c );
assign a = ((~b) & (~c) );
assign a = ( ~b & ~c );
assign a = ( (~b) & (~c) );
assign a = ( ~b & ~c);
assign a = ( (~b) & (~c));