I am getting X as clk_out.Example of the frequency divider.
module freq_by2(input clk,rst,output reg clk_out);
always #(posedge clk)
begin
if (rst==1'b1)
begin
clk_out<=0;
end
else
begin
clk_out<=~clk_out;
end
end
endmodule
module freq_by4(input clk,rst,output reg [1:0] clk_out);
freq_by2 f0(.clk(clk),.rst(rst),.clk_out(clk_out[0]));
freq_by2 f1(.clk(clk_out[0]),.rst(rst),.clk_out(clk_out[1]));
endmodule
Are you sure, you are generating a correct, synchronous rst pulse? That is, synchronous to both clk and clk/2, because your f1 instantiation is synchronous to the clk_out output of f0? Otherwise, you are negating the last value of clk_out inside of freq_by2, which is X initially, unless a correct reset was generated.
the question in it itself is simple. Is it possible to synthesize an Asynchronous counter in Verilog?
More explanation:
So for example, if I have something like the following code, is it possible to synthesize it?
always #(posedge clk) begin
//rst
if(!rst)begin
enable <=0;
end else begin
enable <= 1;
end
end
//action loop
always #(state) begin
case (state)
0:begin
cnt <= cnt
end
1: begin
cnt <= cnt + 1;
next_state <= 1;
end
default: begin
cnt <= cnt;
end
endcase
end
//state loop
always #(next_state, control, enable) begin
if(enable)begin
if(!control) begin
state <= next_state;
end else begin
state <= 0;
end
end
end
Here the general idea is that the counter will go asynchronously while the input control flag is 0, if it is 1 then the counter will be stopped until the input control flag becomes 0 again.
Note: I know I could try and synthesize the code and see what happens. But before that, I would like to know more and see if people have tried it.
Thanks for reading!!
Your action loop can't be properly synthesized.
What it is doing when state == 1?
Your state loop is synthesizable except but you will produce a latch (to store the value when enable == 0). This block is something that could be done using combinational logic '=' instead of non-blocking assignment '<='.
I know this is pushing the good will of the community by presenting my least elaborate work expecting someone to come and save me but I simply have no choice with nothing to lose. I've gone through packets, files, types, flags and boxes the last few weeks but I haven't covered much of recursion. Especially not drawing with recursion. My exam is in roughly one week and I hope this is ample time to repeat and learn simple recursion tricks like that of drawing bowling pins or other patterns:
I I I I I
I I I I
I I I
I I
I
n = 5
The problem I have with recursion is that I don't quite get it. How are you supposed to approach a problem like drawing pins like this using recursion?
The closest I've come is
I I I
I I
I
n = 3
and that's using
THIS CODE NOW WORKS
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Integer_Text_IO; use Ada.Integer_Text_IO;
procedure pyramidchaser is
subtype X_Type is Integer range 0..30;
X: X_Type;
Indent : Integer := 0;
procedure Numbergrabber (X : out Integer) is
begin
Put("Enter a number: ");
Get(X);
Skip_Line;
end Numbergrabber;
procedure PrintSpaces(I : in Integer) is
begin
if I in X_Type then
Put(" ");
else
return;
end if;
PrintSpaces(I - 1);
end Printspaces;
procedure PrintPins(i, n: in Integer) is
begin
if i >= n then
return;
end if;
Put('I');
Put(' ');
PrintPins(i + 1, n);
end Printpins;
function Pins (X, Indent: in Integer) return Integer is
Printed : Integer;
begin
Printed:= 0;
if X > 0 then
PrintSpaces(Indent);
PrintPins(0, X);
New_Line;
Printed := X + Pins(X - 1, Indent + 1);
end if;
return Printed;
end Pins;
Bowlingpins : Integer;
begin
Numbergrabber(X);
Bowlingpins:= Pins(X, Indent);
end pyramidchaser;
but with that I throw in a sum I dont really need just to kick off the recursive part which I dont really know why I do except it seems to be needed to be there. I experimented with code from a completely different assignment, thats why it looks the way it does. I know mod 2 will grant me too many new lines but at least it was an approach to finding heights to the pyramid. I understand the real approach is something similar to N+1 as with each step of the growing pyramid a new line is needed, but I dont know how to implement it.
I don't expect anyone to present a complete code but I hope somebody can clue me in on how to think on the way towards finding a solution.
I can still pass the exam without knowing recursion as its typically 2 assignments where one is and one isnt recursion and you need to pass one or the other, but given that I have some time I thought Id give it a chance.
As always, immensely thankful for anyone fighting the good fight!
Seeing this post gathered some attention Id like to expand the pyramid to one a little more complex:
THE PYRAMID PROBLEM 2
hope someone looks at it. I didnt expect so many good answers, I thought why not throw all I have out there.
Level 1
|=|
Level 2
|===|
||=||
|===|
Level 3
|=====|
||===||
|||=|||
||===||
|=====|
it needs to be figured out recursively. so some way it has to build both upwards and downwards from the center.
To clarify Im studying for an exam and surely so are many others whom would be thankful for code to sink their teeth in. Maybe theres an easy way to apply what Tama built in terms of bowling pin lines in pyramid walls?
Bowling pins:
Printing ----I---- is just: (I'm going to use dashes instead of spaces throughout for readability)
Put_Line (4 * "-" & "I" & 4 * "-");
And printing the whole bowling triangle could be:
with Ada.Strings.Fixed; use Ada.Strings.Fixed;
with Ada.Text_IO; use Ada.Text_IO;
procedure Main is
procedure Print_Bowling_Line (Dashes : Natural; Pins : Positive) is
Middle : constant String := (if Pins = 1 then "I"
else (Pins - 1) * "I-" & "I");
begin
Put_Line (Dashes * "-" & Middle & Dashes * "-");
end Print_Bowling_Line;
begin
Print_Bowling_Line (0, 5);
Print_Bowling_Line (1, 4);
Print_Bowling_Line (2, 3);
Print_Bowling_Line (3, 2);
Print_Bowling_Line (4, 1);
end Main;
Writing the five repeated lines as a loop is quite obvious. For recursion, there are two ways.
Tail recursion
Tail recursion is the more natural 'ask questions, then shoot' approach; first check the parameter for an end-condition, if not, do some things and finally call self.
procedure Tail_Recurse (Pins : Natural) is
begin
if Pins = 0 then
return;
end if;
Print_Bowling_Line (Total_Pins - Pins, Pins);
Tail_Recurse (Pins - 1);
end Tail_Recurse;
Head recursion
Head recursion is what mathematicians love; how do you construct a proof for N? Well, assuming you already have a proof for N-1, you just apply X and you're done. Again, we need to check the end condition before we go looking for a proof for N-1, otherwise we would endlessly recurse and get a stack overflow.
procedure Head_Recurse (Pins : Natural) is
begin
if Pins < Total_Pins then
Head_Recurse (Pins + 1); -- assuming N + 1 proof here
end if;
Print_Bowling_Line (Total_Pins - Pins, Pins);
end Head_Recurse;
For the full code, expand the following snippet:
with Ada.Strings.Fixed; use Ada.Strings.Fixed;
with Ada.Text_IO; use Ada.Text_IO;
procedure Main is
Total_Pins : Natural := 5;
procedure Print_Bowling_Line (Dashes : Natural; Pins : Positive) is
Middle : constant String := (if Pins = 1 then "I"
else (Pins - 1) * "I-" & "I");
begin
Put_Line (Dashes * "-" & Middle & Dashes * "-");
end Print_Bowling_Line;
procedure Tail_Recurse (Pins : Natural) is
begin
if Pins = 0 then
return;
end if;
Print_Bowling_Line (Total_Pins - Pins, Pins);
Tail_Recurse (Pins - 1);
end Tail_Recurse;
procedure Head_Recurse (Pins : Natural) is
begin
if Pins < Total_Pins then
Head_Recurse (Pins + 1); -- assuming N + 1 proof here
end if;
Print_Bowling_Line (Total_Pins - Pins, Pins);
end Head_Recurse;
begin
Total_Pins := 8;
Head_Recurse (1);
end Main;
For simplicity, I don't pass around the second number, that indicates the stopping condition, but rather set it once before running the whole.
I always find it unfortunate to try to learn a technique by applying it where it makes the code more complicated, rather than less. So I want to show you a problem where recursion really does shine. Write a program that prints a maze with exactly one path between every two points in the maze, using the following depth-first-search pseudo code:
start by 'visiting' any field (2,2 in this example)
(recursion starts with this:)
while there are any neighbours that are unvisited,
pick one at random and
connect the current field to that field and
run this procedure for the new field
As you can see in the animation below, this should meander randomly until it gets 'stuck' in the bottom left, after which it backtracks to a node that still has an unvisited neighbour. Finally, when everything is filled, all the function calls that are still active will return because for each node there will be no neighbours left to connect to.
You can use the skeleton code in the snippet below. The answer should only modify the Depth_First_Make_Maze procedure. It need be no longer than 15 lines, calling Get_Unvisited_Neighbours, Is_Empty on the result, Get_Random_Neighbour and Connect (and itself, of course).
with Ada.Strings.Fixed; use Ada.Strings.Fixed;
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Containers; use Ada.Containers;
with Ada.Containers.Vectors;
with Ada.Numerics.Discrete_Random;
procedure Main is
N : Positive := 11; -- should be X*2 + 1 for some X >= 1
type Cell_Status is (Filled, Empty);
Maze : array (1 .. N, 1 .. N) of Cell_Status := (others => (others => Filled));
procedure Print_Maze is
begin
for Y in 1 .. N loop
for X in 1 .. N loop
declare
C : String := (case Maze (X, Y) is
--when Filled => "X", -- for legibility,
when Filled => "█", -- unicode block 0x2588 for fun
when Empty => " ");
begin
Put (C);
end;
end loop;
Put_Line ("");
end loop;
end Print_Maze;
type Cell_Address is record
X : Positive;
Y : Positive;
end record;
procedure Empty (Address : Cell_Address) is
begin
Maze (Address.X, Address.Y) := Empty;
end Empty;
procedure Connect (Address1 : Cell_Address; Address2 : Cell_Address) is
Middle_X : Positive := (Address1.X + Address2.X) / 2;
Middle_Y : Positive := (Address1.Y + Address2.Y) / 2;
begin
Empty (Address1);
Empty (Address2);
Empty ((Middle_X, Middle_Y));
end Connect;
function Cell_At (Address : Cell_Address) return Cell_Status is (Maze (Address.X, Address.Y));
function Left (Address : Cell_Address) return Cell_Address is (Address.X - 2, Address.Y);
function Right (Address : Cell_Address) return Cell_Address is (Address.X + 2, Address.Y);
function Up (Address : Cell_Address) return Cell_Address is (Address.X, Address.Y - 2);
function Down (Address : Cell_Address) return Cell_Address is (Address.X, Address.Y + 2);
type Neighbour_Count is new Integer range 0 .. 4;
package Neighbours_Package is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Cell_Address);
use Neighbours_Package;
function Get_Unvisited_Neighbours (Address : Cell_Address) return Neighbours_Package.Vector is
NeighbourList : Neighbours_Package.Vector;
begin
NeighbourList.Reserve_Capacity (4);
if Address.X >= 4 then
declare
L : Cell_Address := Left (Address);
begin
if Cell_At (L) = Filled then
NeighbourList.Append (L);
end if;
end;
end if;
if Address.Y >= 4 then
declare
U : Cell_Address := Up (Address);
begin
if Cell_At (U) = Filled then
NeighbourList.Append (U);
end if;
end;
end if;
if Address.X <= (N - 3) then
declare
R : Cell_Address := Right (Address);
begin
if Cell_At (R) = Filled then
NeighbourList.Append (R);
end if;
end;
end if;
if Address.Y <= (N - 3) then
declare
D : Cell_Address := Down (Address);
begin
if Cell_At (D) = Filled then
NeighbourList.Append (D);
end if;
end;
end if;
return NeighbourList;
end Get_Unvisited_Neighbours;
package Rnd is new Ada.Numerics.Discrete_Random (Natural);
Gen : Rnd.Generator;
function Get_Random_Neighbour (List : Neighbours_Package.Vector) return Cell_Address is
Random : Natural := Rnd.Random (Gen);
begin
if Is_Empty (List) then
raise Program_Error with "Cannot select any item from an empty list";
end if;
declare
Index : Natural := Random mod Natural (List.Length);
begin
return List (Index);
end;
end Get_Random_Neighbour;
procedure Depth_First_Make_Maze (Address : Cell_Address) is
begin
null;
end Depth_First_Make_Maze;
begin
Rnd.Reset (Gen);
Maze (1, 2) := Empty; -- entrance
Maze (N, N - 1) := Empty; -- exit
Depth_First_Make_Maze ((2, 2));
Print_Maze;
end Main;
To see the answer, expand the following snippet.
procedure Depth_First_Make_Maze (Address : Cell_Address) is
begin
loop
declare
Neighbours : Neighbours_Package.Vector := Get_Unvisited_Neighbours (Address);
begin
exit when Is_Empty (Neighbours);
declare
Next_Node : Cell_Address := Get_Random_Neighbour (Neighbours);
begin
Connect (Address, Next_Node);
Depth_First_Make_Maze (Next_Node);
end;
end;
end loop;
end Depth_First_Make_Maze;
Converting recursion to loop
Consider how a function call works; we take the actual parameters and put them on the call stack along with the function's address. When the function completes, we take those values off the stack again and put back the return value.
We can convert a recursive function by replacing the implicit callstack containing the parameters with an explicit stack. I.e. instead of:
procedure Foo (I : Integer) is
begin
Foo (I + 1);
end Foo;
We would put I onto the stack, and as long as there are values on the stack, peek at the top value and do the body of the Foo procedure using that value. When there is a call to Foo within that body, push the value you would call the procedure with instead, and restart the loop so that we immediately start processing the new value. If there is no call to self in this case, we discard the top value on the stack.
Restructuring a recursive procedure in this way will give you an insight into how it works, especially since pushing on to the stack is now separated from 'calling' that function since you explicitly take an item from the stack and do something with it.
You will need a stack implementation, here is one that is suitable:
Bounded_Stack.ads
generic
max_stack_size : Natural;
type Element_Type is private;
package Bounded_Stack is
type Stack is private;
function Create return Stack;
procedure Push (Onto : in out Stack; Item : Element_Type);
function Pop (From : in out Stack) return Element_Type;
function Top (From : in out Stack) return Element_Type;
procedure Discard (From : in out Stack);
function Is_Empty (S : in Stack) return Boolean;
Stack_Empty_Error : exception;
Stack_Full_Error : exception;
private
type Element_List is array (1 .. max_stack_size) of Element_Type;
type Stack is
record
list : Element_List;
top_index : Natural := 0;
end record;
end Bounded_Stack;
Bounded_Stack.adb
package body Bounded_Stack is
function Create return Stack is
begin
return (top_index => 0, list => <>);
end Create;
procedure Push (Onto : in out Stack; Item : Element_Type) is
begin
if Onto.top_index = max_stack_size then
raise Stack_Full_Error;
end if;
Onto.top_index := Onto.top_index + 1;
Onto.list (Onto.top_index) := Item;
end Push;
function Pop (From : in out Stack) return Element_Type is
Top_Value : Element_Type := Top (From);
begin
From.top_index := From.top_index - 1;
return Top_Value;
end Pop;
function Top (From : in out Stack) return Element_Type is
begin
if From.top_index = 0 then
raise Stack_Empty_Error;
end if;
return From.list (From.top_index);
end Top;
procedure Discard (From : in out Stack) is
begin
if From.top_index = 0 then
raise Stack_Empty_Error;
end if;
From.top_index := From.top_index - 1;
end Discard;
function Is_Empty (S : in Stack) return Boolean is (S.top_index = 0);
end Bounded_Stack;
It can be instantiated with a maximum stack size of Width*Height, since the worst case scenario is when you happen to choose a non-forking path that visits each cell once:
N_As_Cell_Size : Natural := (N - 1) / 2;
package Cell_Stack is new Bounded_Stack(max_stack_size => N_As_Cell_Size * N_As_Cell_Size, Element_Type => Cell_Address);
Take your answer to the previous assignment, and rewrite it without recursion, using the stack above instead.
To see the answer, expand the following snippet.
procedure Depth_First_Make_Maze (Address : Cell_Address) is
Stack : Cell_Stack.Stack := Cell_Stack.Create;
use Cell_Stack;
begin
Push (Stack, Address);
loop
exit when Is_Empty (Stack);
declare
-- this shadows the parameter, which we shouldn't refer to directly anyway
Address : Cell_Address := Top (Stack);
Neighbours : Neighbours_Package.Vector := Get_Unvisited_Neighbours (Address);
begin
if Is_Empty (Neighbours) then
Discard (Stack); -- equivalent to returning from the function in the recursive version
else
declare
Next_Cell : Cell_Address := Get_Random_Neighbour (Neighbours);
begin
Connect (Address, Next_Cell);
Push (Stack, Next_Cell); -- equivalent to calling self in the recursive version
end;
end if;
end;
end loop;
end Depth_First_Make_Maze;
You’re going to print as many lines as there are pins. Each line has an indentation consisting of a number of spaces and a number of pins, each printed as "I " (OK,there's an extra space at the end of the line, but no one's going to see that).
Start off with no leading spaces and the number of pins you were asked to print.
The next line needs one more leading space and one fewer pin (unless, of course, that would mean printing no pins, in which case we're done).
I don't program in Ada (it looks like a strange Pascal to me), but there is an obvious algorithmic problem in your Pins function.
Basically, if you want to print a pyramid whose base is N down to the very bottom where base is 1, you would need to do something like this (sorry for crude pascalization of the code).
procedure PrintPins(i, n: Integer)
begin
if i >= n then return;
Ada.Text_IO.Put('I'); // Print pin
Ada.Text_IO.Put(' '); // Print space
PrintPins(i + 1, n); // Next iteration
end;
function Pins(x, indent: Integer): Integer
printed: Integer;
begin
printed := 0;
if x > 0 then
PrintSpaces(indent); // Print indentation pretty much using the same artificial approach as by printing pins
PrintPins(0, x);
(*Print new line here*)
(*Now go down the next level and print another
row*)
printed := x + Pins(x - 1, indent + 1);
end;
return printed;
end
P.S. You don't need a specific function to count the number of printed pins here. It's just a Gauss sequence sum of the range 1..N, which is given by N(N + 1)/2
A variation of this program is to use both head recursion and tail recursion in the same procedure.
The output of such a program is
Enter the number of rows in the pyramid: 5
I I I I I
I I I I
I I I
I I
I
I
I I
I I I
I I I I
I I I I I
N = 5
Tail recursion produces the upper triangle and head recursion produces the lower triangle.
The way you develop a recursive algorithm is to pretend that you already have a subprogram that does what you want, except for the first bit, and you know, or can figure out, how to do the first bit. Then your algorithm is
Do the first bit
Call the subprogram to do the rest on what remains,
taking into account the effect of the first bit, if necessary
The trick is that "Call the subprogram to do the rest" is a recursive call to the subprogram you're creating.
It's always possible that the subprogram may be called when there's nothing to do, so you have to take that into account:
if Ending Condition then
Do any final actions
return [expression];
end if;
Do the first bit
Call the subprogram to do the rest
And you're done. By repreatedly doing the first bit until the ending condition is True, you end up doing the whole thing.
As an example, the function Get_Line in Ada.Text_IO may be implemented (this is not how it is usually implemented) by thinking, "I know how to get the first Character of the line. If I have a function to return the rest of the line, then I can return the first Character concatenated with the function result." So:
function Get_Line return String is
C : Character;
begin
Get (Item => C);
return C & Get_Line;
end Get_Line;
But what if we're already at the end of a line, so there's no line to get?
function Get_Line return String is
C : Character;
begin
if End_Of_Line then
Skip_Line;
return "";
end if;
Get (Item => C);
return C & Get_Line;
end Get_Line;
For your problem the first bit is printing a row with an indent and a number of pins, and the ending condition is when there are no more rows to print.
For your pyramid problem, this tail recursion scheme doesn't work. You need to do "middle recursion":
if Level = 1 then
Print the line for Level
return
end if
Print the top line for Level
Recurse for Level - 1
Print the bottom line for Level
Verilog case statmenet expression is always true
module test(input clk,
input reset,
output reg[3:0] ledss
);
reg[31:0] dataread;
always #(posedge clk)
begin
case(dataread)
32'b1010101010101010101:ledss='b1010;
endcase
end
endmodule
dont understand why this line executed
32'b1010101010101010101:ledss='b1010;
leds is on same pattern 1010
also after executing this code , leds not on
module test(input clk,
input reset,
output reg[3:0] ledss
);
reg[31:0] dataread;
always #(posedge clk)
begin
if(dataread==32'b1010101010101010101) ledss='b1010;
end
endmodule
but if i execute this,leds is on ,pattern 1010
module test(input clk,
input reset,
output reg[3:0] ledss
);
reg[31:0] dataread;
always #(posedge clk)
begin
case(dataread)
32'b101010101010:
begin
if(dataread==32'b101010101010) ledss='b1010;
end
endcase
end
endmodule
dont understand how case statement works in verilog
I believe you are skipping Verilog simulation and loading your code directly to FPGA.
In simulation. ledss would be X until the patter match is satisfied (both case or if). In the provided code dataread is never assigned so it will be X, therefore in simulation ledss will always be X. In your pastebin links you have dataread driven by a ROM output, so it will have a know output that might match the checking condition.
FPGA synthesizes your RTL and typically goes through some optimization. ledss is not explicitly initialized and has only one possible value if ever assigned. Because of this the optimizer might assume the initial value is don't care, then simplify logic by choosing the initial value to be the same as the only possible value it can be assigned to. Or it might assume the initial value is 0 and keep the logic. For this scenario in general, case tends the follow the former and if tends to follow the latter. Though your code is functionally equivalent, it is uncommon to have a case-statement with only one condition.
I suggest you improve your coding style so your intended behavior is more explicitly understood by the synthesizer and anyone reading your code. Bellow are some suggestions. Remember to assign dataread to a known value. (Note: replaced 32'b1010101010101010101 with the equivalent 32'h0005_5555 for human readability)
always #(posedge clk)
begin
case(dataread)
32'h0005_5555 : ledss <= 4'b1010;
default : ledss <= 4'b1111;
endcase
end
Or equivalent:
always #(posedge clk)
begin
if (dataread == 32'h0005_5555)
ledss <= 4'b1010;
else
ledss <= 4'b1111;
end
If you want ledss to keep the 1010 patter after assignment, then you could do:
always #(posedge clk)
begin
if (reset) begin
ledss <= 4'b1111;
end
case(dataread)
32'h0005_5555 : ledss <= 4'b1010;
endcase
end
Or equivalent:
always #(posedge clk)
begin
if (reset) begin
ledss <= 4'b1111;
end
else if (dataread == 32'h0005_5555) begin
ledss <= 4'b1010;
end
end
The module to be verified is as follows...
The module has an input in1 and an output out1, on alternating clock cycles, out1 is the buffered and inverted value of in1.
I tried coding the checker module using a recursive property.
module check (input in1, out1, clk);
property p1(bit state);
bit nxt;
#(posedge clk)
(1 ,nxt=!state) |-> (out1 == (nxt) ? !in1 : in1) and
nexttime p1(nxt);
endproperty
initial assert property(p1(0)) else $fatal(3, "Failed!");
endmodule
However, running the code at edaplayground throws this error...
RROR VCP2000 "Syntax error. Unexpected token: nexttime[_NEXTTIME]. The
'nexttime' is a SystemVerilog keyword and cannot be used as an
identifier.
I know that this assertion can be done without recursion but I would like to use recursion to write it.
The error says that you have used nexttime which is a systemverilog property keyword in a wrong manner. This operator checks that "if the clock ticks once more, then a shall be true at the next clock tick" in the following code
property p1;
nexttime a;
endproperty
By default, the concurrent assertions shall be checked on every clock pulse, so there is no need of recursion here. Roughly, you can do something like this:
module check (input in1, out1, clk);
bit o_nxt;
function bit update(input bit nxt);
o_nxt = nxt;
return 1;
endfunction
property p1(bit state);
bit nxt;
#(posedge clk)
(1 ,nxt=!state) |-> (out1 == (nxt) ? !in1 : in1) and
update(nxt);
endproperty
initial assert property(p1(o_nxt)) else $fatal(3, "Failed!");
endmodule