How can I delete empty Verilog modules, using sed and xargs?

Question

I have empty modules in my Verilog file (written automatically) which I want to remove upon writing the file.
e.g.

module noneed ;  
endmodule

My plan is to find these lines and remove them from my file, so I came up with this (non-working) pipeline:

grep -P -A 1 -n 'module \S+ \;' myfile.v \
  | perl -p -e 's/(\d+).*/$1/' \
  | grep -v '-' \
  | xargs -0 -i sed '{}d' myfile.v

My line: find empty modules, take their line numbers (and the line after) and feed it to sed for removing.

What have I done wrong here and how can I fix my one-liner to actually remove the lines?

BTW, xargs is cool and everything, but I don't care for a solution not using it.

Edit
here's a test text:

buf syn_1 ( .a ( clkin ) , .vcc ( vcc ) ,  .vss ( vss ) ) ;  
endmodule  
module some_bu11shit_module_nobody_needs_234 ;  
endmodule  
module a_real_module ( clk , inputs , vcc , vss ) ;  
input  clk ;

EDIT2
desired output:

buf syn_1 ( .a ( clkin ) , .vcc ( vcc ) ,  .vss ( vss ) ) ;  
endmodule  
module a_real_module ( clk , inputs , vcc , vss ) ;  
input  clk ;

It seems you already use Perl. Why not to use [Verilog::Parser](http://p3rl.org/Verilog::Parser)? — choroba, Jan 23 '18 at 10:23
never used it (or heard of it till now). in any case, i'm aiming to a one-liner and i don't know how to use CPANs in oneliners — user2141046, Jan 23 '18 at 10:33
Possible duplicate of [sed or awk: delete n lines following a pattern](https://stackoverflow.com/questions/4396974/sed-or-awk-delete-n-lines-following-a-pattern) — Sundeep, Jan 23 '18 at 10:57
see also https://stackoverflow.com/questions/17908555/printing-with-sed-or-awk-a-line-following-a-matching-pattern — Sundeep, Jan 23 '18 at 10:58
@Sundeep - thanks for the links, but sed doesn't support PCRE regex syntax, so I really can't give it a pattern that describes things of the sort of `a_one234five_cell_name` — user2141046, Jan 23 '18 at 11:37
GNU sed supports `\S` and all POSIX tools support `[^[:space:]]` .. and you can use `[a-zA-Z0-9_]` if those are the valid characters allowed — Sundeep, Jan 23 '18 at 11:38
I'm in a tcsh (should be GNU), in any case, neither `sed -e '/module \S+ \;/,+1d'` or `'/module [^[:space:]] \;/,+1d'` works. Thanks for your effort, btw... — user2141046, Jan 23 '18 at 11:42
Can you show the desired output corresponding to your test input? That would help clarify the question. As it is, I don't see why a simple `sed -i -e '/^module noneed/,/endmodule/' *.v` isn't sufficient. — Toby Speight, Jan 23 '18 at 11:53
@Sundeep - getting `sed: invalid option -- 'E'` when using -E @TobySpeight - edited the question to have the desired output as well — user2141046, Jan 23 '18 at 12:13

stevesliva · Answer 1 · 2018-01-24T14:36:10.570

1

sed -e '/\bmodule[^(]*$/,/endmodule/d' myfile.v

For any line with word boundary + module and no open parens, delete until endmodule.

Effectively removes modules with no pins, for typical ways of formatting Verilog.

A robust solution would instead use semicolons as line delimiters.

That said, this is /START/,/END/{ ... } syntax we use with sed quite a lot editing netlists:

sed \
    -e '/\.SUBCKT DELETEME\b/,/\.ENDS/d' \
    -e '/module DELETETHISTOO\b/,/endmodule/d' \
    input-file > output-file

... because even though spice can have line continuations and verilog can have arbitrary line breaks until a semicolon, most netlisting tools can be told to avoid those more difficult cases.

edited Jan 24 '18 at 14:36

answered Jan 24 '18 at 02:50

stevesliva

5,351
1
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your answer seems valid, but it doesn't do the trick. ls -l myfile.v ... 65363912 ... myfile.v sed -e '/module[^(]*$/,/endmodule/d' myfile.v ls -l myfile.v ... 65363912 ... myfile.v – user2141046 Jan 24 '18 at 09:05
Was working worse for me with some broadened testcases. Need either a line start anchor (`^`) or a word boundary (`\b`) before `module` or it matches every endmodule line. Hopefully works for you now. – stevesliva Jan 24 '18 at 14:28

score 0 · Accepted Answer · answered Jan 23 '18 at 12:17

0

with the help of @Sundeep, I came up with this line that does it super quick.

sed -i -re '/module \S+ ;/,+1d' myfile.v

Thanks

answered Jan 23 '18 at 12:17

user2141046

862
2
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How can I delete empty Verilog modules, using sed and xargs?

2 Answers2