Really interesting! I changed the script to:

#!/usr/bin/perl
use strict;
use warnings;

my $a="  456 789 123 456 789  ";
my $count;
$count = $a =~ /\d+/g ;
print "$count\n";

my @a = $a =~ /\d+/g ;
$count = @a;
print "@a\n";
print "$count\n";

$count = () = $a =~ /\d+/g ;
print "$count\n";
[download]

The results are:<\p>

1
789 123 456 789
4
5
[download]

Could you please advise why the 1st digit group is gone the first time i call $a =~ /\d+/g ?

When you use the g modifier, it keeps track of its last match. In the first RE (scalar context), 456 is matched: your count indicates a single match (see code below for actual value). In the second RE (list context), the remaining four sets of digits are matched. You can use pos to reset the last match position.

$ perl -Mstrict -Mwarnings -e '
> my $a="  456 789 123 456 789  ";
> my $count;
> $count = $a =~ /\d+/g ;
> print "$count\n";
> print "${^MATCH}\n";
> pos($a) = 0;
> my @a = $a =~ /\d+/g ;
> $count = @a;
> print "@a\n";
> print "$count\n";
> '
1
456
456 789 123 456 789
5
[download]

${^MATCH} is described in perlvar.

-- Ken

Thanks so much !! It's great to learn more about how g behaves :-)

Hint: look for the phrase "search position" in perlop. Also consider that scalar global matches are typically used in loops, for example

for my $match ($a =~ /\d+/g) {
   print "$match\n"
}
[download]

Really i could not understand but the code

$count = $a =~ /\d+/g ;
vs.
$count = () = $a =~ /\d+/g ;
[download]

is an issue.