Perl script to grep for always block and print the specified lines - perl

Question

I need to fetch the files , read line by line and search for always_ff block.
If always_ff is present, then i need to print the specific lines from inside the found always_ff block
use strict;
use warnings;
for my $file ('*.sv')
{
open my $fh, "<", $file or die $!;
while (<$fh>) {
print scalar <$fh> if /begin/;
}
}
The file has :
always_ff #(negedge clk )
begin
if (!clk)
begin
p1 <= 1'b0;
end
else
begin
p1 <= disable_ff;
end
end
always_ff #(negedge clk)
begin
if (!clk)
begin
p2<= enable;
end
else
begin
p2 <= disable_ff;
end
end
Expected result:
p1 <= disable_ff
p2<= enable
p2 <= disable_ff;
Actual result coming :
if (!clk)
p1 <= disable_ff;

Answer 1

You should probably use the range operator with patterns that match the start and the end of the section. Looking at your input data I would suggest:
start: match "always_ff" plus white space, i.e. /always_ff\s+/
end: empty line, i.e. /^\s*$/
in the block: print lines with assignment, i.e. /<=/
Quote from perlop:
In scalar context, .. returns a boolean value. The operator is bistable, like a flip-flop, ... It is false as long as its left operand is false. Once the left operand is true, the range operator stays true until the right operand is true, AFTER which the range operator becomes false again.
Something like this
#!/usr/bin/perl
use warnings;
use strict;
while (<DATA>) {
if (/always_ff\s+/../^\s*$/) {
print if /<=/;
}
}
exit 0;
__DATA__
always_ff #(negedge clk )
begin
if (!clk)
begin
p1 <= 1'b0;
end
else
begin
p1 <= disable_ff;
end
end
always_ff #(negedge clk)
begin
if (!clk)
begin
p2<= enable;
end
else
begin
p2 <= disable_ff;
end
end
Test output:
$ perl dummy.pl
p1 <= 1'b0;
p1 <= disable_ff;
p2<= enable;
p2 <= disable_ff;
For your use case of reading *.sv files replace <DATA> from my test code with <STDIN> and use the following shell command line:
$ cat *.sv | perl script.pl