"While" effect on regex?

[lightoverhead]

Dear Monks,

I have questions regarding the "while" effect on regular expression matching process. Below is the code:

use v5.14;
while(<>) {
while( m/
        \b
        (\w\S+)
        (
            \s+
            \1
        ) +
        \b
        /xi  #if add modifier "g", it works fine.
)
{
say "dup word '$1' at paragraph $.";
}
}

My questions are:

1. This code will produce the first matched duplicate word infinitely. It seems that the second "while" evaluate the condition of regex matching unlimited times. Why doese this happen? Does the matching only happen once?

2. But if I add "g" modifier to the regex matching, it will accurately produce all the matched strings once and stop. Why does this happen?

3. What is the effect of second "while"? when a "while" used in a regex pattern matching statement, what is the behavior of "while",or how many times will "while" evaluate regex conditions, especially with or without modifier "g"?

Thank you.

[roboticus]

lightoverhead:

Without the 'g' modifier, the m/.../ operator simply reports whether the string matches or not. Unless you change the string, it will *always* match or *always* fail. If you read perlre, it'll refer you to perlretut. In there it says:

Global matching

The final two modifiers "//g" and "//c" concern multiple matches. The modifier "//g" stands for global matching and allows the matching operator to match within a string as many times as possible. In scalar context, successive invocations against a string will have `"//g" jump from match to match, keeping track of position in the string as it goes along. You can get or set the position with the "pos()" function.

The relevent bit is the italicized bit. It's telling you that with the g operator, the match remembers where it left off, and the next time you match with the g operator, it will proceed from that point.

Long story short: that's what the 'g' modifier does!

...roboticus

When your only tool is a hammer, all problems look like your thumb.

[lightoverhead]

roboticus

Your answer is very clear! Thanks a lot!

[7stud]

1. This code will produce the first matched duplicate word infinitely. It seems that the second "while" evaluate the condition of regex matching unlimited times. Why doese this happen? Does the matching only happen once?

In scalar context, the match operator returns 0 (no match) or 1 (match). The conditional for a while loop is a boolean context, and a boolean context is considered a scalar context because a boolean context demands either a number or a string, which are both scalars. Anything else is converted to a number or string, and then the number or string is evaluated as being true or false.

Any function (or operator) in your code is replaced by the function's return value. So in your while loop conditional, the match operator will return 1 if it finds a match, and therefore your while loop becomes:

while (1) {
    ...
}

Also, because you have a capture group in your regex, the match operator sets $1. The end result is this loop:

while (1) {
  say $1;
}

The m// operator only looks for the first match and then quits:

use strict;
use warnings;
use 5.012;

if ('aXaYaZ' =~ /(a[XYZ])/ ) {
    say $1;
}

--output:--
aX

my @matches =  'aXaYaZ' =~ /(a[XYZ])/g; 
say "@matches";

--output:--
aX

The /g flag which stands for 'global matching' changes that behavior.

[Anonymous Monk]

why is the second while loop needed ?

while(<>) {
while( m/\b
        (\w\S+)
....

and not just ?

while(<>) {
if ( m/\b
        (\w\S+)
...

[CountZero]

The first while loop reads in the data to be matched. The second while loop goes through all the matches (if the g modifier is added).