RandomPkg: How to set weight for range

Why OSVVM™?ForumsOSVVMRandomPkg: How to set weight for range

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This topic has 3 voices, contains 4 replies, and was last updated by Avatar of Eilert Backhus Eilert Backhus 359 days ago.

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June 22, 2017 at 12:04 #1387
Avatar of Nikolay
Nikolay

Hi,

Is it possible to reduce next code:

     A <= RndA.DistValSlv(
       ((0,5),
        (1,1),(2,1),(3,1),(4,1),
        (5,1),(6,1),(7,1),(8,1),(9,1),
        (10,1),(11,1),(12,1),(13,1),(14,1),
        (15,5)), A'length);

I need to set Weight for range. For example in SV it looks like this:

a dist {0 := 5, [1:4] /= 5, [5:9] /=5, [10:14] /= 5, 15 := 5}

Thanks,

Nikolai

June 26, 2017 at 01:51 #1389
Avatar of Eilert Backhus
Eilert Backhus

Hi Nicolai,

the argument type of DistValSlv is an array with a record of two integers.

I wonder wether the range notation can work here too.

It wold look somewhat like this:

A <= RndA.DistValSlv(
       ((0,5),
        (1 to 14,1),
        (15,5)), A'length);

If this works not, maybe a function with a case statement can help.

In the function you then write sth. like this:

for position in 0 to 15 loop — 15 might come from some ‘lengh calculation
  case position is
    when 0 =>  ReturnVal(position,5);
    when 1 to 14 =>ReturnVal(position,1);
    when 15 =>ReturnVal(position,5);
  end case;
end loop
return ReturnVal;

Unless there isn’t some better solution, this might at least be a useful  workaround for the moment.

(edit:) But actualy that won’t reduce the code much, unless you have a really big array of random constraints.

Have a nice simulation

  Eilert

June 28, 2017 at 19:38 #1393
Avatar of Jim Lewis
Jim Lewis

Hi Nikolai,

There is a simpler form of DistValSlv that returns values in the range of its argument, which is 0 to 15 if the argument is a literal or aggregate:

     A <= RndA.DistSlv((5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5), A'length) ;

If you prefer, you can also use named association:

     A <= RndA.DistSlv((0=>5, 1=>1, 2=>1, 3=>1, 4=>1, 5=>1, 6=>1,
                        7=>1, 8=>1, 9=>1, 10=>1, 11=>1, 12=>1,
                        13=>1, 14=>1, 15=>5), A'length) ;

With named association, you can also do the following:

     A <= RndA.DistSlv((0=>5, 1 to 14 =>1, 15=>5), A'length) ;


You can also do this with a coverage model:

     shared variable ACov : CovPType ;
     . . .
     process
     begin
       ACov.AddBins(5, GenBin(0)) ;

       ACov.AddBins(1, GenBin(1,14)) ;
       ACov.AddBins(5, GenBin(15)) ;

       . . .
       A <= ACov.RandCovPoint ;
       ACov.ICoverLast ;

Jim


June 28, 2017 at 22:58 #1396
Avatar of Nikolay
Nikolay

Hi Eilert, Jim,

Thank you for the answers. They are very useful for me.

Nikolai

June 28, 2017 at 23:11 #1398
Avatar of Eilert Backhus
Eilert Backhus

Hi Jim,

I hoped that there’s some solution using the range notation like you showed:

With named association, you can also do the following:

     A <= RndA.DistSlv((0=>5, 1 to 14 =>1, 15=>5), A'length) ;

I just didn’t expect it to be so straight simple, since the defined type is self defined.
You mentioned integer_vectors somewhere else. Are these defined internally in a similar way as DistValSlv is? Does this kind of assignment work with integer_vectors too?

Kind regards
  Eilert

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