... what would be the scope of \G. Supposes in next print statement I use a different string, then also \G would will make pattern match to start from a position where it last matched in string one?

Each individual string has an independent "position of end of last successful match" attribute that is returned by the pos built-in. The  \G regex operator (enabled by the  /g modifier) accesses this attribute of a string being matched to assert that matching in that string is continuing where previous matching in that string by any  m//g match left off. 

c:\@Work\Perl\monks>perl -wMstrict -le
"my $s1 = 'foobarfeefiefoefum';
 $s1 =~ /foo/g;
 ;;
 my $s2 = '123456789';
 $s2 =~ /6/g;
 ;;
 print qq{A: pos in \$s1 '$s1' after successful match == }, pos $s1;
 print qq{B: pos in \$s2 '$s2' after successful match == }, pos $s2;
 ;;
 $s1 =~ /foe/g;
 print qq{C: pos in \$s1 '$s1' after successful match == }, pos $s1;
 print qq{D: pos in \$s2 '$s2' still == }, pos $s2;
"
A: pos in $s1 'foobarfeefiefoefum' after successful match == 3
B: pos in $s2 '123456789' after successful match == 6
C: pos in $s1 'foobarfeefiefoefum' after successful match == 15
D: pos in $s2 '123456789' still == 6

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What would have happened if the second match against the  $s1 string had been  /\Gfoe/g instead? Or  /\Gbar/g instead? Try it and see! (See also the documentation concerning the effect of the  /c modifier in conjunction with  /g in a  m//gc match.)

(Incidentally, what if  $test_string in your example code was  '12xxx345' and the match was  /(\d)/g (no \G) instead? What if it was  /\G(\d)/g as originally?)

Give a man a fish:  <%-{-{-{-<

In reply to Re^7: Regular expression by AnomalousMonk
in thread Regular expression by pravakta

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