RC4 - someone help pleas!!

A

Ahmed Samieh

Guest
Hi all,

i'm tring to implement RC4 algorithm on fpga, i wrote the vhdl code
and simulation work fine too
but when i want to synthesize the errors and warnings pops up in my
face :(

LIBRARY ieee;
USE ieee.std_logic_1164.all;
USE ieee.numeric_std.all;
--
ENTITY RC4 IS
GENERIC (d_width : natural := 8;
a_width : natural := 8);
PORT(clk : IN std_logic;
data_en : IN std_logic;
seed : IN std_logic_vector(63 DOWNTO 0) :=
X"0369CF258BE147AD";
data_out : OUT std_logic_vector(d_width-1 DOWNTO 0));
END ENTITY RC4;
--
ARCHITECTURE Arch OF RC4 IS
SUBTYPE byte IS natural RANGE 0 TO 2**d_width-1;
TYPE ram IS ARRAY (natural RANGE <>) OF byte;
SIGNAL x_s,y_s : byte;
SIGNAL rst_1_s,rst_2_s : std_logic := '0';
BEGIN
rise : PROCESS(clk,data_en,seed,rst_1_s,rst_2_s)
VARIABLE temp_1_v,temp_2_v,temp_3_v : byte;
VARIABLE state_v : ram(0 TO 2**a_width-1);
VARIABLE j_v,x_v,y_v : byte;
VARIABLE seed_v : ram(0 TO 7);
-- ATTRIBUTE logic_block : boolean;
-- ATTRIBUTE logic_block OF state_v: VARIABLE IS true;
BEGIN
IF (rst_1_s = rst_2_s) THEN
FOR i IN 7 DOWNTO 0 LOOP
seed_v(i) := to_integer( unsigned( seed((i+1)*8-1 DOWNTO
i*8) ) );
END LOOP;
FOR i IN 0 TO 2**a_width-1 LOOP
state_v(i) := i;
END LOOP;
FOR i IN 0 TO 2**a_width-1 LOOP
temp_1_v := seed_v(i mod 8);
temp_2_v := state_v(i);
j_v := (temp_1_v + temp_2_v + j_v) mod 2**a_width;
temp_3_v := state_v(j_v);
state_v(i) := temp_3_v;
state_v(j_v) := temp_2_v;
END LOOP;
x_s <= 0;
y_s <= 0;
rst_1_s <= NOT(rst_2_s);
data_out <= (OTHERS => '0');
ELSIF rising_edge(clk) THEN
IF (data_en = '1') THEN
x_v := (x_s + 1) mod 2**a_width;
y_v := (y_s + state_v(x_v)) mod 2**a_width;
temp_1_v := state_v(x_v);
temp_2_v := state_v(y_v);
state_v(x_v) := temp_2_v;
state_v(y_v) := temp_1_v;
temp_3_v := state_v((temp_1_v + temp_2_v) mod 2**a_width);
data_out <= std_logic_vector(to_unsigned(temp_3_v, d_width));
x_s <= x_v;
y_s <= y_v;
END IF;
END IF;
END PROCESS rise;
new_frame_reset : PROCESS(data_en,rst_1_s)
BEGIN
IF rising_edge(data_en) THEN
rst_2_s <= rst_1_s;
END IF;
END PROCESS new_frame_reset;
END ARCHITECTURE Arch;

thanx,

Ahmed Samieh
 
On Jun 9, 11:29 pm, Ahmed Samieh <Asm4...@gmail.com> wrote:
Hi all,
Click to expand...
1st problem come from

FOR i IN 0 TO 2**a_width-1 LOOP
state_v(i) := i;
END LOOP;
Click to expand...
where state_v is 256*8 ram (array of 256 bytes)

but i don't understand why i got such a warning

Warning: 5783: Module WiFi_MAC.Test(Arch){generic map (d_width => 8
a_width => 8)}, Net rst_1_s: This signal has multiple drivers. This
may lead to simulation mismatch.

???

any suggestions?

thanx,

Ahmed Samieh
 
"Ahmed Samieh" <Asm4pic@gmail.com> wrote in message
news:1181436802.653489.302550@m36g2000hse.googlegroups.com...
On Jun 9, 11:29 pm, Ahmed Samieh <Asm4...@gmail.com> wrote:
Hi all,

1st problem come from

FOR i IN 0 TO 2**a_width-1 LOOP
state_v(i) := i;
END LOOP;

where state_v is 256*8 ram (array of 256 bytes)

but i don't understand why i got such a warning

Warning: 5783: Module WiFi_MAC.Test(Arch){generic map (d_width => 8
a_width => 8)}, Net rst_1_s: This signal has multiple drivers. This
may lead to simulation mismatch.
Click to expand...
The warning is complaining about the signal rst_1_s. You are assigning to it
twice, and that leads to "multiple drivers".

-Michael.
 
On Jun 11, 2:00 pm, "Michael Jřrgensen" <i...@ukendt.dk> wrote:
"Ahmed Samieh" <Asm4...@gmail.com> wrote in message

news:1181436802.653489.302550@m36g2000hse.googlegroups.com...





On Jun 9, 11:29 pm, Ahmed Samieh <Asm4...@gmail.com> wrote:
Hi all,

1st problem come from

FOR i IN 0 TO 2**a_width-1 LOOP
state_v(i) := i;
END LOOP;

where state_v is 256*8 ram (array of 256 bytes)

but i don't understand why i got such a warning

Warning: 5783: Module WiFi_MAC.Test(Arch){generic map (d_width => 8
a_width => 8)}, Net rst_1_s: This signal has multiple drivers. This
may lead to simulation mismatch.

The warning is complaining about the signal rst_1_s. You are assigning to it
twice, and that leads to "multiple drivers".

-Michael.- Hide quoted text -

- Show quoted text -
Click to expand...
thanx Michael,

but if you talk a look in the code you will find only one driver for
rst_1_s
only
rst_1_s <= NOT(rst_2_s);
Click to expand...
so what is wrong with this code ?

Ahmed Samieh
 
On Fri, 15 Jun 2007 15:54:12 -0700, Ahmed Samieh <Asm4pic@gmail.com>
wrote:

Warning: 5783: Module WiFi_MAC.Test(Arch){generic map (d_width => 8
a_width => 8)}, Net rst_1_s: This signal has multiple drivers. This
may lead to simulation mismatch.

The warning is complaining about the signal rst_1_s. You are assigning to it
twice, and that leads to "multiple drivers".

thanx Michael,

but if you talk a look in the code you will find only one driver for
rst_1_s

rst_1_s <= NOT(rst_2_s);
and
Click to expand...
SIGNAL rst_1_s,rst_2_s : std_logic := '0';

Make that two.

- Brian
 
On Jun 16, 4:22 pm, Brian Drummond <brian_drumm...@btconnect.com>
wrote:
rst_1_s <= NOT(rst_2_s);

and
SIGNAL rst_1_s,rst_2_s : std_logic := '0';

Make that two.

- Brian
Click to expand...
it is only an initial value to the signal not signal driver, so there
is no repot about rst_2_s

Ahmed Samieh
 
On Sun, 17 Jun 2007 04:47:03 -0700, Ahmed Samieh <Asm4pic@gmail.com>
wrote:

On Jun 16, 4:22 pm, Brian Drummond <brian_drumm...@btconnect.com
wrote:
rst_1_s <= NOT(rst_2_s);

and
SIGNAL rst_1_s,rst_2_s : std_logic := '0';

Make that two.

- Brian

it is only an initial value to the signal not signal driver, so there
is no repot about rst_2_s
Click to expand...
You're right, it's just a default value; this doesn't create a driver.
A signal always has a default value; you've just specified an explicit
one here, rather than using the 'implicit default' (in Verilog,
though, initialisation is just a shorthand for a continuous
assignment, so this *would* be a problem).

I think your problem is simply that your code is unsynthesisable, and
your tool is just trying (unsuccessfully) to tell you that. rst_1_s is
part of a combinatorial feedback loop (look at 'rst_1_s = rst_2_s' and
'rst_1_s <= not(rst_2_s'), *and* you're trying to give it a default
value. You need to sketch out what hardware you're trying to build,
and then try to code up that hardware, using some combination of
clocked and combinatorial processes.

Evan
 
On Mon, 18 Jun 2007 09:57:18 +0100, Evan Lavelle <nospam@nospam.com>
wrote:


(in Verilog,
though, initialisation is just a shorthand for a continuous
assignment, so this *would* be a problem).
Click to expand...
Technically, that should have been 'a blocking assignment in an
initial construct', rather than 'a continuous assignment'.

Evan
 
On Jun 18, 11:57 am, Evan Lavelle <nos...@nospam.com> wrote:
I think your problem is simply that your code is unsynthesisable, and
your tool is just trying (unsuccessfully) to tell you that. rst_1_s is
part of a combinatorial feedback loop (look at 'rst_1_s = rst_2_s' and
'rst_1_s <= not(rst_2_s'), *and* you're trying to give it a default
value. You need to sketch out what hardware you're trying to build,
and then try to code up that hardware, using some combination of
clocked and combinatorial processes.

Evan
Click to expand...
thanx Evan,

ok..the problem in the code and the point which can't be done within
the code is :
i need to access 256 location using loop - only during one clock !!!

Ahmed Samieh
 
Hey Ahmed,

I think you have several problems in your code. But I don't agree that
writing to 256 location in one cycle would be a problem, provided you don't
want a RAM. Obviously if you want to write to a RAM you need to do it line
by line etc.

Ok, to cut a long story short. The problem appears to come from two things:

firstly you make signals rst_1_s and rst_2_s and init them to '0'. be
cautious here, this won't simulate the same as it runs...

secondly, you've some crazy combinational logic with the two reset signals
and data_en. At the very least this _must_ make a latch for rst_1_s
somewhere.

Have a look at what you're trying to do, and (as others have already
suggested) try drawing out the circuit first by hand.

I also think you're cheating the simulator by including data_en in your
sensitivity list for 'rise' process.

to summarise:
-when data_en has a rising_edge state rst_2_s becomes the same as rst_1_s
-immediately these are equal, 'rise' executes and sets them inequal through
a combinational path.
-so you have a tiny ~ns pulse driving the first part of the 'rise' process
-BUT you have a clock AND data_en = '1' comparison also in the 'rise'
process ...

This is very confusing - how exactly are you trying to do this?? Have
another look through and post some more code when you make some progress :)

Ben
 
On Mon, 18 Jun 2007 07:23:46 -0700, Ahmed Samieh <Asm4pic@gmail.com>
wrote:

On Jun 18, 11:57 am, Evan Lavelle <nos...@nospam.com> wrote:
I think your problem is simply that your code is unsynthesisable, and
your tool is just trying (unsuccessfully) to tell you that. rst_1_s is
part of a combinatorial feedback loop (look at 'rst_1_s = rst_2_s' and
'rst_1_s <= not(rst_2_s'), *and* you're trying to give it a default
value. You need to sketch out what hardware you're trying to build,
and then try to code up that hardware, using some combination of
clocked and combinatorial processes.

Evan

thanx Evan,

ok..the problem in the code and the point which can't be done within
the code is :
i need to access 256 location using loop - only during one clock !!!
Click to expand...
I haven't looked at your code in detail, but at first sight the loops
look synthesisable (but will turn into a lot of hardware). The problem
is more likely to be the combinatorial feedback loop I mentioned
above. What hardware are you trying to create with rst_1_s/rst_2_s in
your process "reset" branch? Can you draw a schematic for it?

Evan
 
thanx Benjamin, thanx Even

ok the problem is that it is a software algorithm basically
http://en.wikipedia.org/wiki/RC4_(cipher)

so i have to run the initialization function during rise edge of
data_en

for i from 0 to 255
S := i
endfor
j := 0
for i from 0 to 255
j := (j + S + key[i mod keylength]) mod 256
swap(S,S[j])
endfor
i := 0
j := 0

then for each clock and during data_en = '1' i have to run the prga
function

i := (i + 1) mod 256
j := (j + S) mod 256
swap(S,S[j])
output S[(S + S[j]) mod 256]

that is the algorithm.

the code in vhdl without any modification , and not free of errors :

LIBRARY ieee;
USE ieee.std_logic_1164.all;
USE ieee.numeric_std.all;
--
ENTITY RC4 IS
GENERIC (d_width : natural := 8;
a_width : natural := 8);
PORT(clk : IN std_logic;
data_en : IN std_logic;
seed : IN std_logic_vector(63 DOWNTO 0);
data_out : OUT std_logic_vector(d_width-1 DOWNTO 0));
END ENTITY RC4;
--
ARCHITECTURE Arch OF RC4 IS
SUBTYPE byte IS natural RANGE 0 TO 2**d_width-1;
TYPE ram IS ARRAY (natural RANGE <>) OF byte;
SIGNAL seed_s : ram(0 TO 7);
SIGNAL x_s,y_s : byte;
BEGIN
seed_convert : PROCESS(seed)
BEGIN
FOR i IN 7 DOWNTO 0 LOOP
seed_s(i) <= to_integer(unsigned(seed((i+1)*8-1 DOWNTO i*8)));
END LOOP;
END PROCESS seed_convert;
--
main : PROCESS(clk,data_en,seed_s)
VARIABLE temp_1_v,temp_2_v,temp_3_v : byte;
VARIABLE state_v : ram(0 TO 2**a_width-1);
VARIABLE j_v,x_v,y_v : byte;
BEGIN
IF rising_edge(data_en) THEN
FOR i IN 0 TO 2**a_width-1 LOOP
state_v(i) := i;
END LOOP;
j_v := 0;
FOR i IN 0 TO 2**a_width-1 LOOP
j_v := (seed_s(i mod 8) + state_v(i) + j_v) mod 2**a_width;
temp_1_v := state_v(i); -- swap(state_v(i),state(j_v))
temp_2_v := state_v(j_v); -- swap(state_v(i),state(j_v))
state_v(i) := temp_1_v; -- swap(state_v(i),state(j_v))
state_v(j_v) := temp_1_v; -- swap(state_v(i),state(j_v))
END LOOP;
x_s <= 0;
y_s <= 0;
data_out <= (OTHERS => '0');
ELSIF rising_edge(clk) THEN
IF (data_en = '1') THEN
x_v := (x_s + 1) mod 2**a_width;
y_v := (y_s + state_v(x_v)) mod 2**a_width;
temp_1_v := state_v(x_v);
temp_2_v := state_v(y_v);
state_v(x_v) := temp_2_v;
state_v(y_v) := temp_1_v;
temp_3_v := state_v((temp_1_v + temp_2_v) mod 2**a_width);
data_out <= std_logic_vector(to_unsigned(temp_3_v, d_width));
x_s <= x_v;
y_s <= y_v;
END IF;
END IF;
END PROCESS main;
END ARCHITECTURE Arch;

the algorithm is :

- if rising_edge(data_en) then initialize state
- elsif rising_edge(clk) and data_en = '1' then prga

* initialize state
- state is 256 byte from 0 to 255
- set state(i) = i;
then
using seed, swap state -the algorithm above-
- now it is ready to prga

* prga
for each clock output 1 byte and swap 2 bytes in state

problems....
1 - error to drive process with 2 clocks - rising_edge(data_en) and
rising_edge(clk) -
2 - i'm tring to access 2 memory locations at the same time for booth
read and write, and repeat this 256 times only in halve clock (start
from riseing_edge(data_en) and before rising_edge(clock) - where
data_en changed with falling_edge(clk) ) .... complex ?? :(

Ahmed Samieh
 
On Jun 19, 2:50 pm, Ahmed Samieh <Asm4...@gmail.com> wrote:
state_v(i) := temp_1_v; -- swap(state_v(i),state(j_v))
Click to expand...
sorry,
state_v(i) := temp_2_v; -- swap(state_v(i),state(j_v))

Ahmed Samieh
 
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