Can someone explain how clogb2 function works ?

[mr_vasanth]

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Code Verilog - [expand]
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function integer clogb2;
input[31:0] value;
for(clogb2=0; value>0; clogb2=clogb2+1)
value = value>>1;
endfunction

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Assume the input value is 1.
I just could not understand how the recursive function call works and when clogb2=clogb2+1 gets executed.

Can someone explain this ?

 

[dave_59]

This is not a recursive function call. In Verilog, the name of the function is used as variable representing the return value. There was no return statement in Verilog. It has been added by SystemVerilog.

Also, Verilog-2005 added $clog2, so there is no need to write this function yourself.

 

[mr_vasanth]

Thanks dave.

I need some more info. What will be the value of clogb2 when the loop exits ?
Assume value = 1 is passed as an input.

1. First clogb2 will be initialized to 0
2. Checks the condition value > 0, condition passed
3. Right shift value by 1, value becomes 0 post shift
4. Increment clogb2 = clogb2 + 1
5. Check the condition value > 0, condition failed and loop exits.

When value 1 is passed as an input, logb2 value should be 0 to my knowledge. But if I go by the steps listed above the value we get is 1.
Please correct me if the above steps listed by me has any flaw.

 

[dave_59]

20.8.1 Integer math functions (1800-2012)
The system function $clog2 shall return the ceiling of the log base 2 of the argument (the log rounded up to an integer value). The argument can be an integer or an arbitrary sized vector value. The argument shall be treated as an unsigned value, and an argument value of 0 shall produce a result of 0.
This system function can be used to compute the minimum address width necessary to address a memory of a given size or the minimum vector width necessary to represent a given number of states.

 

[mr_vasanth]

dave,

Thank you so much for your time and explanation. I am aware of what you have written in your second post.

My only confusion is how this for loop works ? how does it return a value of 0 when the input argument is 1 ?
for(clogb2=0; value>0; clogb2=clogb2+1)
value = value>>1;
endfunction

Would be great if you could explain only how does this for loop works ?

 

[dave_59]

You answered this question in #3. When the input is value is 1, clog2b is 1 because the for-loop iterates once.

 

[mr_vasanth]

But the logb2 value of 1 is supposed to be 0 right ?
Does it mean the above code is wrong ?
Then what is the right way to code ?
Please help.

 

[dave_59]

That might be true for logb2, but this is clogb2. This is returning the minimum number of bits needed to represent a value.

 

[mr_vasanth]

That makes sense. But this triggers another question.

We need at least one bit to represent the value 1. I agree to that.
In the same way, Don't we need at least one bit to represent the value 0 ?

I believe this function clogb2 cannot be used (or invalid) when the input value is 0.

 

[dave_59]

From a synthesis perspective, you don't need any bits to represent the a variable whose only value is zero; it becomes a constant. And constant 0 can always be optimized away.

 

[mr_vasanth]

is clogb2 synthesizable ?

 

[mr_vasanth]

There is an error in my code Dave. The corrected one you can find below.
I have earlier missed the statement value = value-1; That was causing all my confusion.
Now the function produces exact logb2 value for all the value, except 0.

function integer clogb2;
  input integer value;
  begin
    [B]value = value-1;[/B]
    for (clogb2=0; value>0; clogb2=clogb2+1)
      value = value>>1;
  end
endfunction