Is it possible to increment/initialize a variable inside 'generate' in verilog?

Question

I have a doubt in the generate, my code is:

parameter m=1;

generate 
 for(i=0; i<m; i=i+1) :loopstart
 begin
  statements;
 end
endgenerate

Inside this loop, m should be 2^0, 2^1, 2^2, and so on. Since exponentiation is not supported, I thought of initializing m and then multiplying it by 2 on each iteration.

I have a few questions:

Is it possible to use m << 1 inside the generate in some way (since this is the same as multiplying by 2)? If I do that, it results in an error.

I referred to Samir Palnitkar's book, which says that always statement works inside a generate, so I tried:

always @(m)
 m <= m*2; // (or m << 1)

This doesn't work. I realize it can't be done because m is a parameter and not a variable.

If what I think is right, it can't be done with genvar either, since a genvar can't be initialized.

Is there an alternative?

Morgan · Accepted Answer · 2013-07-19 07:07:43Z

The question seem to have been created because exponentials are not supported. They are :

2**0 => 1
2**1 => 2
2**2 => 4

NB:

always @(m)
 m <= m*2; // (or m << 1)

This is wrong for a few reasons

Do not use <= in combinatorial blocks
In the above expression m defines itself and re-triggers the loop.

I would also avoid named sensitivity lists and use always @* to avoid hardware simulation mismatches from incomplete sensitivity lists.

parameters and localparams are for defining constants, if they are not constant use something else like a logic or integer type.

logic [31:0] m = 0;
logic [31:0] power_two_m;
always @* begin
  power_two_m = 2**m;
  power_two_m = 1 << m;
end

Dominik Honnef · Accepted Answer · 2012-11-18 12:34:50Z

0

Instead i++ use i=i+1

And.. forget C.

edited Nov 18, 2012 at 12:34

Dominik Honnef

18.5k7 gold badges43 silver badges43 bronze badges

answered Apr 13, 2011 at 9:46

user705733

11 bronze badge

4 Comments

Nandhini Over a year ago

on no :( i typed i++ here by mistake. i have entered it correctly in the program..

Nandhini Over a year ago

I have solved it.. i took m in another loop above i loop and did m=m*2, thanks all :)

Andy Over a year ago

@Nandhini please post an answer describing how you resolved the problem

Nandhini Over a year ago

@ Andy : I had to do m=2^j for j=0,1,2 and so on inside a for loop inside 'generate' such that m=1,2,4,.... Instead what i did was that i removed j from the loop and just took the variable m inside the for loop and wrote for(m=1;condition;m=m*2). this way it works.

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