On Jan 8, 12:21 pm, frank.w.w...@gmail.com (Soldier) wrote:
> Hi,
> I came across these two pieces of codes, why would the "local $i=$i+1"
> be backtracking-safe?
>
> $_ = 'lothlorien';
> m/ (?{ $i = 0 }) # Set $i to 0
> (. (?{ $i++ }) )* # Update $i, even after
> backtracking
> lori # Forces a backtrack
> /x;
>
> $_ = 'lothlorien';
> m/ (?{ $i = 0 })
> (. (?{ local $i = $i + 1; }) )* # Update $i, backtracking-safe.
> lori
> (?{ $result = $i }) # Copy to non-localized location.
> /x;
I don't know how the internals but, adding a bit of
debug exposes what happens IIUC:
m/ ...
(. (?{ local $i = $i + 1; warn "\$i=$i\n";}) )*
...
/x;
warn "\$i=$i \$result=$result\n"
m/ ...
(. (?{ $i++; warn "\$i=$i\n";}) )*
...
/x;
warn "\$i=$i \$result=$result\n"
The output below suggests both regexes find the initial
match at pos. 4 within the string; pos. 4 is saved; but
* greedily continues to bump along failing with each bump
until the end of the string is reached. Then with 'local'
the first and only match at pos. 4 gets restored. Without
the 'local' though, $i runs up to the end of the string at
pos. 10.
$i=1
$i=2
...
$i=10
$i=0 $result=4
$i=1
$i=2
...
$i=10
$i=10 $result=10
--
Charles DeRykus
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