minimal matching to end of string in regular expressions

ISAI student has asked for the wisdom of the Perl Monks concerning the following question:

Hello all. I am trying to get minimal matching to the end of the string. I will explain by an example. The code is:

perl -e '$a ="Inst1 cell1 ( .Q ( nets ) , .T (nety)  ) ;\n"; $a=~s/\).
+*?$//; print $a;'
[download]

The output is:

Inst1 cell1 ( .Q ( nets
[download]

I want it to be:

"Inst1 cell1 ( .Q ( nets ) , .T (nety)  "
[download]

Now, I know of ways to do it, but not in 1 regexp

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Re: minimal matching to end of string in regular expressions by MVS (Monk) on Mar 04, 2012 at 06:26 UTC
If you want to be sure that you're matching the last right parenthesis: `$a =~ s/\)[^)]*$//;`	[reply] [d/l]
Re: minimal matching to end of string in regular expressions by jwkrahn (Abbot) on Mar 04, 2012 at 06:32 UTC
`$ perl -e '$a ="Inst1 cell1 ( .Q ( nets ) , .T (nety) ) ;\n"; $a=~s/( +.\))./$1/; print $a;' Inst1 cell1 ( .Q ( nets ) , .T (nety) )` [download]	[reply] [d/l]
Re: minimal matching to end of string in regular expressions by JavaFan (Canon) on Mar 04, 2012 at 11:43 UTC
In general, if you want to find the last pattern in a string: `/.PAT/s;` [download] So, in your case: `s/(.)\)./$1/s;` [download] You can also combine this with the \K construct: `s/.\K\).*//s` [download] BTW, your question makes it unclear to me whether you want to trailing newline to be removed. My code does, and your "want it to be" does not include a newline -- but your original solution does.	[reply] [d/l] [select]
Re: minimal matching to end of string in regular expressions by toolic (Bishop) on Mar 04, 2012 at 13:23 UTC
That looks like Verilog code, which can be parsed with Verilog::Netlist.	[reply]
Re: minimal matching to end of string in regular expressions by Anonymous Monk on Mar 04, 2012 at 18:20 UTC
Better clear than clever.	[reply]