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Perl: Getting back reference from s modifier


 
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# 1  
Old 03-17-2008
Perl: Getting back reference from s modifier

My input text has the following pattens:
Code:
  func_a(3,
         4,
         5);

I want to replace it with this:
Code:
   
  func_b(3,
         4,
         5,
         6);

I'm trying the following expression, but it does not work:
Code:
perl -p -e "s/func_a\((.*)?\);/func_b(\1,\n6)/s" < file |more

Any ideas? Thanks.
# 2  
Old 03-17-2008
use $1 and $6 instead of \1 and \6, see if that helps. Use single-quotes around the code unless you are on windows.
# 3  
Old 03-18-2008
Thanks, but that does not work. By the way, I'm not using \6, but rather \n6.
# 4  
Old 03-18-2008
Use $1 instead of \1

Modify the command as below ...
perl -p -e "s/func_a\((.*)?\);/func_b\($1,\n6\);/s" < file |more
# 5  
Old 03-18-2008
Yes, I used $1 instead of \1. I think it makes no difference.
When I have patterns that don't have new lines in them, such as:
Code:
  func_a(3);

Doing
Code:
perl -p -e "s/func_a\((.*)?\);/func_b($1,\n6)/s" < file |more

works as expected.
# 6  
Old 03-18-2008
The problem is that with perl -p, you are only reading and examining a line at a time. Thus the pattern match is not being applied to anything which straddles a line boundary, even if you have the /s modifier. You can fix that by reading all lines at once, with perl -0777 or something equivalent.

See also the Perl FAQ, which has a detailed discussion of this topic.
era
# 7  
Old 03-18-2008
perl should have given me an error/warning because the s modifier isn't applicable to -p.
Anyway, here's what I tried, but this time I get nothing outputted:
Code:
perl -e -0777 "s/func_a\((.*)?\);/func_b(\1,\n6)/s" < file |more

and
Code:
perl -e "BEGIN {$/=undef} s/func_a\((.*)?\);/func_b(\1,\n6)/s"


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