Test vector for only MSB being set.

[Niv (KP)]

I have a function which accepts an unsigned vector of unconstrained
length, and assigns it to the variable "work".

One of the operations it performs depends on the value being
"1000......0000"
that is all '0's axcept the msb being set to a '1' , for any length
vector

I currently test for this as follows:

ELSIF ((work(work'HIGH) = '1')) AND
((work(work'HIGH - 1 DOWNTO work'LOW)) = (work(work'HIGH - 1
DOWNTO work'LOW)'RANGE => '0')) THEN
-- operation here


This compiles OK, but is there a neater way, it does seem a bit long
winded.

Regards, Kev P.

 

[Jonathan Bromley]

On Tue, 3 Feb 2009 05:45:54 -0800 (PST), "Niv (KP)"
<kev.parsons@mbda-systems.com> wrote:

I have a function which accepts an unsigned vector of unconstrained
length, and assigns it to the variable "work".

One of the operations it performs depends on the value being
"1000......0000"
that is all '0's axcept the msb being set to a '1' , for any length
vector

Awkward. Is this any help?

if vec = (1 => '1', 2 to vec'length => '0') then ...

My instinct tells me it is better to write a function:

function msb_only(n: positive) return std_logic_vector is
variable it: std_logic_vector(1 to n);
begin
it := (1 => '1', 2 to n => '0');
return it;
end;
...
if vec = msb_only(vec'length) then ...
--
Jonathan Bromley, Consultant

DOULOS - Developing Design Know-how
VHDL * Verilog * SystemC * e * Perl * Tcl/Tk * Project Services

Doulos Ltd., 22 Market Place, Ringwood, BH24 1AW, UK
jonathan.bromley@MYCOMPANY.com
http://www.MYCOMPANY.com

The contents of this message may contain personal views which
are not the views of Doulos Ltd., unless specifically stated.

 

[Tricky]

On 3 Feb, 13:45, "Niv (KP)" <kev.pars...@mbda-systems.com> wrote:

I have a function which accepts an unsigned vector of unconstrained
length, and assigns it to the variable "work".

One of the operations it performs depends on the value being
"1000......0000"
that is all '0's axcept the msb being set to a '1' , for any length
vector

I currently test for this as follows:

ELSIF ((work(work'HIGH) = '1')) AND
        ((work(work'HIGH - 1 DOWNTO work'LOW)) = (work(work'HIGH - 1
DOWNTO work'LOW)'RANGE => '0'))      THEN
  -- operation here

This compiles OK, but is there a neater way, it does seem a bit long
winded.

Regards, Kev P.

how about changing:
((work(work'HIGH - 1 DOWNTO work'LOW)) = (work(work'HIGH - 1 DOWNTO
work'LOW)'RANGE => '0'))

to

to_integer( ((work(work'HIGH - 1 DOWNTO work'LOW)) ) = 0;

Thats about as compact as it can get. I can only suggest setting up
aliases for the MSB and lower bits, so it reads:

alias MSB : std_logic is work(work'high);
alias LSBs : unsigned(work'HIGH - 1 DOWNTO work'LOW ) is work
(work'HIGH - 1 DOWNTO work'LOW);

elsif MSB = '1' and to_integer(LSBs) = 0 then

This is just reorganisation of the "wind".

 

[Niv (KP)]

On 3 Feb, 14:17, Jonathan Bromley <jonathan.brom...@MYCOMPANY.com>
wrote:

On Tue, 3 Feb 2009 05:45:54 -0800 (PST), "Niv (KP)"

kev.pars...@mbda-systems.com> wrote:
I have a function which accepts an unsigned vector of unconstrained
length, and assigns it to the variable "work".

One of the operations it performs depends on the value being
"1000......0000"
that is all '0's axcept the msb being set to a '1' , for any length
vector

Awkward.  Is this any help?

  if vec = (1 => '1', 2 to vec'length => '0') then ...

My instinct tells me it is better to write a function:

  function msb_only(n: positive) return std_logic_vector is
    variable it: std_logic_vector(1 to n);
  begin
    it := (1 => '1', 2 to n => '0');
    return it;
  end;
  ...
    if vec = msb_only(vec'length) then ...
--
Jonathan Bromley, Consultant

DOULOS - Developing Design Know-how
VHDL * Verilog * SystemC * e * Perl * Tcl/Tk * Project Services

Doulos Ltd., 22 Market Place, Ringwood, BH24 1AW, UK
jonathan.brom...@MYCOMPANY.comhttp://www.MYCOMPANY.com

The contents of this message may contain personal views which
are not the views of Doulos Ltd., unless specifically stated.

The "TO_INTEGER" looks promising, but I've changed it as follows (for
my unique case);

....
ELSIF TO_INTEGER(work) = 2**(work'HIGH) THEN
.....

Kev P.