[SOLVED] Search and replace block of text

sharky · 03-24-2010, 08:20 PM

I thought this would be easy but I can't get it. Oh well.

The goal is to replace this block of text;

Quote:

always begin
if(V(gnd) < 0.01) #1 gnd_d = 1'b0; else #1 gnd_d = 1'b1;
if(V(vdd) > 0.50) #1 vdd_d = 1'b1; else #1 vdd_d = 1'b0;
end

With this;

Quote:

always @(above(V(vdd) - 0.5)) vdd_d=1'b1;
always @(above(0.5 - V(vdd))) vdd_d=1'b0;
always @(above(V(gnd) - 0.1)) gnd_d=1'b1;
always @(above(0.1 - V(gnd))) gnd_d=1'b0;

With sed I can easily extract the unwanted block but I have no idea how to replace it with the desired block.

Manually is out of the question because there are 622 blocks to replace.

grail · 03-24-2010, 08:26 PM

Well, how about you show your extraction and we can see about how to implement?

Curious if your final product has a typo error or not as 0.50 translated to 0.5 but 0.01
somehow translated to 0.1??
Also, not sure if intended, but why is the final output in the reverse order of the original?
Is this required? (ie gnd is in first line of input but last two lines of output)

sharky · 03-25-2010, 10:13 AM

Quote:

Originally Posted by grail

Well, how about you show your extraction and we can see about how to implement?

Curious if your final product has a typo error or not as 0.50 translated to 0.5 but 0.01
somehow translated to 0.1??
Also, not sure if intended, but why is the final output in the reverse order of the original?
Is this required? (ie gnd is in first line of input but last two lines of output)

The replaced text is not a munge of the original. It is simply what is required as a replacement. The extraction is a simple command line sed statement: sed '/always/,/end/p' which removes the first block from the file. That part is easy but it does nothing to replace the new block.

Through BFAI (Brute Force And Ignorance) I've made some progress.

1st, I set the whole block of text to a variable;

Code:

set XVAR = "always @(above(V(vdd) - 0.5)) vdd_d=1'b1;\n   always @(above(0.5 - V(vdd))) vdd_d=1'b0;\n   always @(above(V(gnd) - 0.1)) gnd_d=1'b1;\n   always @(above(0.1 - V(gnd))) gnd_d=1'b0;"

Then a cat and pipe the file like this;

Code:

cat stdcell.vams | sed "s/always begin/$XVAR/" | grep -v 'if(V'

The result is this;

Quote:

always @(above(V(vdd) - 0.5)) vdd_d=1'b1;
always @(above(0.5 - V(vdd))) vdd_d=1'b0;
always @(above(V(gnd) - 0.1)) gnd_d=1'b1;
always @(above(0.1 - V(gnd))) gnd_d=1'b0;
end

Now all I have to do is get rid of 'end' without losing things like 'endif', 'endspecify', 'endmodule', etc...

sharky · 03-25-2010, 10:31 AM

Quote:

Originally Posted by sharky

The replaced text is not a munge of the original. It is simply what is required as a replacement. The extraction is a simple command line sed statement: sed '/always/,/end/p' which removes the first block from the file. That part is easy but it does nothing to replace the new block.

Through BFAI (Brute Force And Ignorance) I've made some progress.

1st, I set the whole block of text to a variable;

Code:

set XVAR = "always @(above(V(vdd) - 0.5)) vdd_d=1'b1;\n   always @(above(0.5 - V(vdd))) vdd_d=1'b0;\n   always @(above(V(gnd) - 0.1)) gnd_d=1'b1;\n   always @(above(0.1 - V(gnd))) gnd_d=1'b0;"

Then a cat and pipe the file like this;

Code:

cat stdcell.vams | sed "s/always begin/$XVAR/" | grep -v 'if(V'

The result is this;

Now all I have to do is get rid of 'end' without losing things like 'endif', 'endspecify', 'endmodule', etc...

Got it!

This is definitely BFAI

After setting the replacement text to a variable I can do this;

Code:

cat stdcells.vams | sed "s/always begin/$XVAR/" | grep -v 'if(V' | grep -wv end

It's ugly but it works.

Sergei Steshenko · 03-25-2010, 10:43 AM

Quote:

Originally Posted by sharky

...
Manually is out of the question because there are 622 blocks to replace.

No, it isn't. You've been waiting for an answer longer that it would take to replace it manually.

...

Regarding your question - I would do it in Perl; regarding 'end' vs 'endif' - there is a notion of word boundary.

If you are interested in doing this with Perl, start from running on your machine

perldoc perlretut
.

...

Nice to see Verilog in this forum

.

ghostdog74 · 03-25-2010, 11:04 AM

use awk

Code:

awk 'BEGIN{
    RS="end"
    s=s"always @(above(V(vdd) - 0.5)) vdd_d=1\047b1;\n"
    s=s"always @(above(0.5 - V(vdd))) vdd_d=1\047b0;\n"
    s=s"always @(above(V(gnd) - 0.1)) gnd_d=1\047b1;\n"
    s=s"always @(above(0.1 - V(gnd))) gnd_d=1\047b0;"
}
/always begin/{gsub(/always begin.*/,s)}1' file

Sergei Steshenko · 03-25-2010, 11:14 AM

Quote:

Originally Posted by ghostdog74

use awk

Code:

awk 'BEGIN{
    RS="end"
    s=s"always @(above(V(vdd) - 0.5)) vdd_d=1\047b1;\n"
    s=s"always @(above(0.5 - V(vdd))) vdd_d=1\047b0;\n"
    s=s"always @(above(V(gnd) - 0.1)) gnd_d=1\047b1;\n"
    s=s"always @(above(0.1 - V(gnd))) gnd_d=1\047b0;"
}
/always begin/{gsub(/always begin.*/,s)}1' file

I think the OP needs to extract first, say,

Code:

(V(gnd) < 0.01)
(V(vdd) > 0.50)

from, say

Code:

if(V(gnd) < 0.01) #1 gnd_d = 1'b0; else #1 gnd_d = 1'b1;
if(V(vdd) > 0.50) #1 vdd_d = 1'b1; else #1 vdd_d = 1'b0;

and then to generate the corresponding 'always' statements. I.e. in "scientific" terms the OP needs Verilog to Verilog translator.

Does your 'awk' code know how to extract the needed expressions in parenthesis ?

sharky · 03-25-2010, 07:10 PM

Quote:

Originally Posted by Sergei Steshenko

No, it isn't. You've been waiting for an answer longer that it would take to replace it manually.

...

Regarding your question - I would do it in Perl; regarding 'end' vs 'endif' - there is a notion of word boundary.

If you are interested in doing this with Perl, start from running on your machine

perldoc perlretut
.

...

Nice to see Verilog in this forum

.

You must be a serious speed demon with cut n paste.

Sergei Steshenko · 03-25-2010, 07:43 PM

Quote:

Originally Posted by sharky

You must be a serious speed demon with cut n paste.

One minute per cut and paste means 662 minutes for the whole job, i.e. 11 hours to do the job manually. Which just proves my point.

Or you think 1 minute per such cut and paste is too little ?

grail · 03-26-2010, 01:10 AM

So I know you said:

Quote:

The replaced text is not a munge of the original.

But I couldn't help myself (been having some fun with awk {although ghostdog might shoot this down in flames})

Code:

awk 'BEGIN{FS="[ \t]*#1[ \t]*"; start="always @(above"}
     $0~FS{gsub(/if|else/,"");
     gsub(/<|>/,"-");
     tmp=gensub(/(V.*) [<|>] (.*)\)/, "\\2 - \\1","g",$1);
     print start $1")", $2;
     print start tmp")", $3}' file

sharky · 03-26-2010, 11:49 AM

Quote:

Originally Posted by Sergei Steshenko

One minute per cut and paste means 662 minutes for the whole job, i.e. 11 hours to do the job manually. Which just proves my point.

Or you think 1 minute per such cut and paste is too little ?

It may have seemed like I spent a lot of time on this because of how long it took me to come back to the forum. But I didn't work all that time on nothing but this script. Each cut n paste would likely take less than a minute. However, I can't ignore the rest of my job working on one fix. As is, the total time I spent on this was nowhere near 8 - 10 hours; More like one hour.

Besides that, the 622 modules was for one set of stdcells on one process. I've already found another set that needs the exact same fix and it contains 503 modules and there could be more. But now I have a solution and it only takes a few seconds and the issue is solved.

Thanks,