note
moritz
<p>It's a good guess, because there is a the fact that $1 points into the original string, and can change when the original string changes. But it's not the source of the confusion here.</p>
<p>You can see that by changing the original code to use <c>m/(\d+)/g</c> instead of <c>s/(\d+)//</c>, thus not modifying the original string at all:
<code>
sub R {
printf qq{before: \$_ is '$_'};
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'
</code>
<p>Let me repeat what I wrote earlier, but hopefully a bit clearer this time: <b>There is only one variable $1</b>. The first call to m/()/ or s/()// creates the dynamic variable $1, and all subsequent calls <b>modify the existing variable $1</b>. 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.</p>
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