+ Post New Thread
Results 1 to 4 of 4

13th June 2017, 20:32 #1
 Join Date
 Jun 2017
 Posts
 1
 Helped
 0 / 0
 Points
 17
 Level
 1
binary dividing: restoring method
Hello,
i have to understand how the restoring method works and found a very helpful source code:
Code:  restoring.vhd   Implements a sequential version of restoring division algorithm  section 3.2 Loopunrolling and digitserial processing  x = xn1 xn2 ... x0: natural  y = yn1 yn2 ... y0: natural  condition: x < y  quotient q = 0.q1 ... qp: nonnegative fractional  remainder r = rn1 rn2 ... r0: natural  x = (0.q1 q2 ... qp)·y + (r/y)·2p with r < y   LIBRARY IEEE; USE IEEE.STD_LOGIC_1164.ALL; USE IEEE.STD_LOGIC_ARITH.ALL; USE IEEE.STD_LOGIC_UNSIGNED.ALL; ENTITY restoring IS GENERIC(n: NATURAL:=8; p: NATURAL:= 6); PORT( x, y: IN STD_LOGIC_VECTOR(n1 DOWNTO 0); clk, reset, start:IN STD_LOGIC; quotient: OUT STD_LOGIC_VECTOR(1 TO p); remainder: OUT STD_LOGIC_VECTOR(n1 DOWNTO 0); done: OUT STD_LOGIC ); END restoring; ARCHITECTURE circuit OF restoring IS SIGNAL r, next_r: STD_LOGIC_VECTOR(n1 DOWNTO 0); SIGNAL long_y, two_r, dif: STD_LOGIC_VECTOR(n DOWNTO 0); SIGNAL load, update: STD_LOGIC; SIGNAL q: STD_LOGIC_VECTOR(1 TO p); SUBTYPE index IS NATURAL RANGE 0 TO p1; SIGNAL count: index; TYPE state IS RANGE 0 TO 3; SIGNAL current_state: state; BEGIN long_y <= '0'&y; two_r <= r&'0'; dif <= two_r  long_y; WITH dif(n) SELECT next_r <= dif(n1 DOWNTO 0) WHEN '0', two_r(n1 DOWNTO 0) WHEN OTHERS; remainder_register: PROCESS(clk) BEGIN IF clk'EVENT AND clk = '1' THEN IF load = '1' THEN r <= x; ELSIF update = '1' THEN r <= next_r; END IF; END IF; END PROCESS; remainder <= r; quotient_register: PROCESS(clk) BEGIN IF clk'EVENT AND clk = '1' THEN IF load = '1' THEN q <= (others => '0'); ELSIF update = '1' THEN q <= q(2 TO p)&NOT(dif(n)); END IF; END IF; END PROCESS; quotient <= q; counter: PROCESS(clk) BEGIN IF clk'EVENT and clk = '1' THEN IF load = '1' THEN count <= 0; ELSIF update = '1' THEN count <= (count+1) MOD p; END IF; END IF; END PROCESS; next_state: PROCESS(clk) BEGIN IF reset = '1' THEN current_state <= 0; ELSIF clk'EVENT AND clk = '1' THEN CASE current_state IS WHEN 0 => IF start = '0' THEN current_state <= 1; END IF; WHEN 1 => IF start = '1' THEN current_state <= 2; END IF; WHEN 2 => current_state <= 3; WHEN 3 => IF count = p1 THEN current_state <= 0; END IF; END CASE; END IF; END PROCESS; output_function: PROCESS(clk, current_state) BEGIN CASE current_state IS WHEN 0 TO 1 => load <= '0'; update <= '0'; done <= '1'; WHEN 2 => load <= '1'; update <= '0'; done <= '0'; WHEN 3 => load <= '0'; update <= '1'; done <= '0'; END CASE; END PROCESS; END circuit;
I started a simulation with modelsim, set the clock, set x and y.
'quotient' have to hold the result, but it is "empty".
Can anybody help me to run the simulation and make a division?
Thanks

13th June 2017, 20:32

14th June 2017, 01:36 #2
 Join Date
 Sep 2013
 Location
 USA
 Posts
 6,551
 Helped
 1584 / 1584
 Points
 28,611
 Level
 41
Re: binary dividing: restoring method
Did you set start to 1? You need that to kick off the FSM.
Note, the output_function process should not have clk in the sensitivity list, that is a potential gotcha as synthesis ignores the sensitivity list (it only looks at the state) even though simulation will enter the process each time the clock toggles (both rising and falling). With all their PhDs and MS degrees you would think they would know that.

14th June 2017, 01:36

14th June 2017, 09:32 #3
 Join Date
 Jun 2010
 Posts
 6,567
 Helped
 1914 / 1914
 Points
 35,892
 Level
 46
Re: binary dividing: restoring method
Its the problem with free code off the internet. Almost always written by hobbiests who dont really understand the language, or come anywhere near decent coding practice (and never document it).
Also note  the "next_state" process is missing reset from it's sensitivity list too! luckily, they get away with it as the initial value for current_state is 0 anyway (and who uses integers for state machines?  are you deliberately avoiding all the synthesis optimisation?).
To The OP: Why not share your testbench also?
   Updated   
I also note this beauty:
count <= (count+1) MOD p
Using a divider unnecessarily (potentially). Normal people would write:
Code:if count = p1 then count = 0 else count <= count + 1; end if;

14th June 2017, 09:32

14th June 2017, 22:22 #4
 Join Date
 Sep 2013
 Location
 USA
 Posts
 6,551
 Helped
 1584 / 1584
 Points
 28,611
 Level
 41
Re: binary dividing: restoring method
You must not have checked the site where this code came from (the site is written by a bunch of PhDs and MS degree holders in electrical engineering). I now understand why we see so many students with equally poor understanding of HDL languages and such poor coding practices.
+ Post New Thread
Please login