Hi monks,

I want to use the back reference in s///g, but seems I failed with code below: 

        $in =~ s/\s+/\n/g;
        $in =~ s/[a-zA-Z]\n[a-zA-Z]/$1\ $2/g;

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First I cut the character (non-English) by space, then I join them back if I cut them wrongly (e.g. space between 2 English alphabet). I found the regular expression in line 2 failed - the space is located correctly, but the character before and after were missing and replaced by null (so I see character disappeared) - then I know the back reference here is not working. Did I failed to call the back reference just because of my stupid syntax error?

In reply to Back reference in s///g ? by Anonymous Monk

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