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Re^3: 'Dynamic scoping' of capture variables ($1, $2, etc.)

by moritz (Cardinal)
on Dec 18, 2012 at 14:22 UTC ( #1009374=note: print w/ replies, xml ) Need Help??


in reply to Re^2: 'Dynamic scoping' of capture variables ($1, $2, etc.)
in thread 'Dynamic scoping' of capture variables ($1, $2, etc.)

It's a good guess, because there is a the fact that $1 points into the original string, and can change when the original string changes. But it's not the source of the confusion here.

You can see that by changing the original code to use m/(\d+)/g instead of s/(\d+)//, thus not modifying the original string at all:

sub R { printf qq{before: \$_ is '$_'}; printf qq{ \$1 is %s \n}, defined($1) ? qq{'$1'} : 'undefined'; m/(\d+)/g ? $1 + R() : 0; printf qq{after: \$_ is '$_'}; printf qq{ \$1 is %s \n}, defined($1) ? qq{'$1'} : 'undefined'; } $_ = 'a81b2d34c1'; R(); __END__ before: $_ is 'a81b2d34c1' $1 is undefined before: $_ is 'a81b2d34c1' $1 is '81' before: $_ is 'a81b2d34c1' $1 is '2' before: $_ is 'a81b2d34c1' $1 is '34' before: $_ is 'a81b2d34c1' $1 is '1' after: $_ is 'a81b2d34c1' $1 is '1' after: $_ is 'a81b2d34c1' $1 is '1' after: $_ is 'a81b2d34c1' $1 is '1' after: $_ is 'a81b2d34c1' $1 is '1' after: $_ is 'a81b2d34c1' $1 is '1'

Let me repeat what I wrote earlier, but hopefully a bit clearer this time: There is only one variable $1. The first call to m/()/ or s/()// creates the dynamic variable $1, and all subsequent calls modify the existing variable $1. Since there is no mechanism for resetting $1 to a previous value, you can see the last value of $1 in all stack frames that have access to it.


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Re^4: 'Dynamic scoping' of capture variables ($1, $2, etc.)
by nvivek (Vicar) on Dec 19, 2012 at 05:13 UTC
    Thanks for your explanation Moritz. I meant $1 value got changed similar to $_ value for better understanding. I know that $_ value getting changed due to substitute operator replaces matched digits to null string. I have a doubt in your example, how $1 value getting changed every recursive call. As per your code, your m// expression will match first digits ( 55 ) at all time as per my understanding. Of course, when I checked the code, it gives $1 value as 55, 666, 7777 and 1 respectively. Kindly give me some detail about it.
      As per your code, your m// expression will match first digits ( 55 ) at all time as per my understanding

      No, because I used m/../g in scalar context, which remembers the previous match position in pos, and then always returns the next match during subsequent calls.

      Maybe you are more familiar with it in this idiom:

      use 5.010; $_ = 'a123b45c6'; while (m/(\d+)/g) { say $1; }
        Thanks for your explanation. I got it.

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