i'm trying to do a 8 states of 3-bit running LED. The LED will move forward when the input x is 1 andmove backward when x is 0. with following conditions.
i've trying this for hours now and i don't know what's wrong with it. i'm learning vhdl right now so pleaseeee go easy on me. it says it's syntax error but i don't know where
library IEEE;
use IEEE.std_logic_1164.all;
entity Quiz10 is
port(
x : in std_logic;
output : out std_logic_vector(6 downto 0)
);
end Quiz10;
architecture arch1 of Quiz10 is
signal digit : integer range 0 to 9 := 0;
begin
process (x,digit)
begin
if x = '1' then
digit <=digit+1;
elsif x = '0' then
digit <=digit-1;
elsif (digit>=9)
digit<=0;
end if;
end process;
output <="1111110" when digit = 0 else
"0110000" when digit = 1 else
"1101101" when digit = 2 else
"1111001" when digit = 3 else
"0110011" when digit = 4 else
"1011011" when digit = 5 else
"1011111" when digit = 6 else
"1110000" when digit = 7 else
"1111111" when digit = 8 else
"1111011" when digit = 9 else
"0000000";
-- Your VHDL code defining the model goes here
end arch1;
I just tried your code with Modelsim and the only syntax error was, as #Tricky pointed out, the missing then after the second elsif.
Related
i'm going to learn C++ at the very beginning and struggling with some challenges from university.
The task was to calculate the cross sum and to use modulo and divided operators only.
I have the solution below, but do not understand the mechanism..
Maybe anyone could provide some advice, or help to understand, whats going on.
I tried to figure out how the modulo operator works, and go through the code step by step, but still dont understand why theres need of the while statement.
#include <iostream>
using namespace std;
int main()
{
int input;
int crossSum = 0;
cout << "Number please: " << endl;
cin >> input;
while (input != 0)
{
crossSum = crossSum + input % 10;
input = input / 10;
}
cout << crossSum << endl;
system ("pause");
return 0;
}
Lets say my input number is 27. cross sum is 9
frist step: crossSum = crossSum + (input'27' % 10 ) // 0 + (modulo10 of 27 = 7) = 7
next step: input = input '27' / 10 // (27 / 10) = 2.7; Integer=2 ?
how to bring them together, and what does the while loop do? Thanks for help.
Just in case you're not sure:
The modulo operator, or %, divides the number to its left by the number to its right (its operands), and gives the remainder. As an example, 49 % 5 = 4.
Anyway,
The while loop takes a conditional statement, and will do the code in the following brackets over and over until that statement becomes false. In your code, while the input is not equal to zero, do some stuff.
To bring all of this together, every loop, you modulo your input by 10 - this will always return the last digit of a given Base-10 number. You add this onto a running sum (crossSum), and then divide the number by 10, basically moving the digits over by one space. The while loop makes sure that you do this until the number is done - for example, if the input is 104323959134, it has to loop 12 times until it's got all of the digits.
It seems that you are adding the digits present in the input number. Let's go through it with the help of an example, let input = 154.
Iteration1
crossSum= 0 + 154%10 = 4
Input = 154/10= 15
Iteration2
crossSum = 4 + 15%10 = 9
Input = 15/10 = 1
Iteration3
crossSum = 9 + 1%10 = 10
Input = 1/10 = 0
Now the while loop will not be executed since input = 0. Keep a habit of dry running through your code.
#include <iostream>
using namespace std;
int main()
{
int input;
int crossSum = 0;
cout << "Number please: " << endl;
cin >> input;
while (input != 0) // while your input is not 0
{
// means that when you have 123 and want to have the crosssum
// you first add 3 then 2 then 1
// mod 10 just gives you the most right digit
// example: 123 % 10 => 3
// 541 % 10 => 1 etc.
// crosssum means: crosssum(123) = 1 + 2 + 3
// so you need a mechanism to extract each digit
crossSum = crossSum + input % 10; // you add the LAST digit to your crosssum
// to make the number smaller (or move all digits one to the right)
// you divide it by 10 at some point the number will be 0 and the iteration
// will stop then.
input = input / 10;
}
cout << crossSum << endl;
system ("pause");
return 0;
}
but still dont understand why theres need of the while statement
Actually, there isn't need (in literal sense) for, number of digits being representable is limited.
Lets consider signed char instead of int: maximum number gets 127 then (8-bit char provided). So you could do:
crossSum = number % 10 + number / 10 % 10 + number / 100;
Same for int, but as that number is larger, you'd need 10 summands (32-bit int provided)... And: You'd always calculate the 10 summands, even for number 1, where actually all nine upper summands are equal to 0 anyway.
The while loop simplifies the matter: As long as there are yet digits left, the number is unequal to 0, so you continue, and as soon as no digits are left (number == 0), you stop iteration:
123 -> 12 -> 1 -> 0 // iteration stops, even if data type is able
^ ^ ^ // to store more digits
Marked digits form the summands for the cross sum.
Be aware that integer division always drops the decimal places, wheras modulo operation delivers the remainder, just as in your very first math lessons in school:
7 / 3 = 2, remainder 1
So % 10 will give you exactly the last (base 10) digit (the least significant one), and / 10 will drop this digit afterwards, to go on with next digit in next iteration.
You even could calculate the cross sum according to different bases (e. g. 16; base 2 would give you the number of 1-bits in binary representation).
Loop is used when we want to repeat some statements until a condition is true.
In your program, the following statements are repeated till the input becomes 0.
Retrieve the last digit of the input. (int digit = input % 10;)
Add the above retrieved digit to crosssum. (crosssum = crosssum + digit;)
Remove the last digit from the input. (input = input / 10;)
The above statements are repeated till the input becomes zero by repeatedly dividing it by 10. And all the digits in input are added to crosssum.
Hence, the variable crosssum is the sum of the digits of the variable input.
I am writing code in Hackerrank. And recently the problem said, convert decimal to base 2 and then count the max consecutive 1's in the binary number. And first I come with following solution. It works fine. But I do not understand the counting part of it, even though I wrote it.
The code is
int main(){
int n,ind=0, count=0, mmax=0;
char bin[100];
cin >> n;
while(n){
if(n%2==0) {
bin[ind]='0';
n = n / 2;
ind = ind + 1;
}
else if(n%2==1) {
bin[ind]='1';
n = n / 2;
ind = ind + 1;
}
}
for(int i=0; i<=(ind-1); i++){
if(bin[i] == '1' && bin[i+1] == '1'){
count++;
if(mmax < count)
mmax = count;
}
else
count=0;
}
cout << mmax + 1 << endl;
return 0;
}
In the above code, I guess that variable mmax will give me the max consecutive number of 1's but it gives me value that has (max consecutive - 1), So I just wrote like that and submitted the code. But I am curious about. why it is working that way. I am little bit of confused the way that code works like this.
Thanks
Lets say you have this binary sequence:
11110
Your code will compare starting from the first and second:
|11|110 1 && 1 -> max = 1
1|11|10 1 && 1 -> max = 2
11|11|0 1 && 1 -> max = 3
111|10| 1 && 0 -> max = 3
you can see, that although there are 4 1's you only do 3 comparisons, so your max will always be -1 of the actual max. You can fix this by adding mmax += 1 after your for loop.
Just a little bit of trace using small example will show why.
First, lets say there is only 1 '1' in your array.
Since you require both the current position and your next position to be '1', you will always get 0 for this case.
Let's say I have "11111". At the first '1', since next position is also '1', you increment count once. This repeats until 4th '1' and you increment your count 4 times in total so far. When you reach 5th '1', your next position is not '1', thus your count stops at 4.
In general, your method is like counting gaps between fingers, given 5 fingers, you get 4 gaps.
Side note: your code will fail for the case when there is no '1' in your array.
I can't understand how to count number of 1's in binary representation.
I have my code, and I hope someone can explain it for me.
Code:
int count (int x)
{
int nr=0;
while(x != 0)
{
nr+=x%2;
x/=2;
}
return nr;
}
Why while ? For example if i have 1011, it wouldn't stop at 0?
Why nr += x%2 ?
Why x/=2 ?!
First:
nr += x % 2;
Imagine x in binary:
...1001101
The Modulo operator returns the remainder from a / b.
Now the last bit of x is either a 0, in which case 2 will always go into x with 0 remainder, or a 1, in which case it returns a 1.
As you can see x % 2 will return (if the last bit is a one) a one, thus incrementing nr by one, or not, in which case nr is unchanged.
x /= 2;
This divides x by two, and because it is a integer, drops the remainder. What this means is is the binary was
....10
It will find out how many times 2 would go into it, in this case 1. It effectively drops the last digit of the binary number because in base 2 (binary) the number of times 2 goes into a number is just the same as 'shifting' everything down a space (This is a poor explanation, please ask if you need elaboration). This effectively 'iterates' through the binary number, allowing the line about to check the next bit.
This will iterate until the binary is just 1 and then half that, drop the remainder and x will equal 0,
while (x != 0)
in which case exit the loop, you have checked every bit.
Also:
'count`is possibly not the most descriptive name for a function, consider naming it something more descriptive of its purpose.
nr will always be a integer greater or equal to zero, so you should probably have the return type unsigned int
int count (int x)
{
int nr=0;
while(x != 0)
{
nr+=x%2;
x/=2;
}
return nr;
}
This program basically gives the numbers of set bits in a given integer.
For instance, lets start with the example integer 11 ( binary representation - 1011).
First flow will enter the while loop and check for the number, if it is equal to zero.
while(11 != 0)
Since 11 is not equal to zero it enter the while loop and nr is assigned the value 1 (11%2 = 1).nr += 11%2;
Then it executes the second line inside the loop (x = x/2). This line of code assigns the value 5 (11/2 = 5 ) to x.
Once done with the body of the while loop, it then again checks if x ie 5 is equal to zero.
while( 5 != 0).
Since it is not the case,the flow goes inside the while loop for the second time and nr is assigned the value 2 ( 1+ 5%2).
After that the value of x is divided by 2 (x/2, 5/2 = 2 )and it assigns 2 to x.
Similarly in the next loop, while (2 != 0 ), nr adds (2 + 2%2), since 2%2 is 0, value of nr remains 2 and value of x is decreased to 1 (2/2) in the next line.
1 is not eqaul to 0 so it enters the while loop for the third time.
In the third execution of the while loop nr value is increased to 3 (2 + 1%2).
After that value of x is reduced to 0 ( x = 1/2 which is 0).
Since it fails the check (while x != 0), the flow comes out of the loop.
At the end the value of nr (Which is the number of bits set in a given integer) is returned to the calling function.
Best way to understand the flow of a program is executing the program through a debugger. I strongly suggest you to execute the program once through a debugger.It will help you to understand the flow completely.
So I have this function that takes in an integer. But It doesn't work and I suspect that the if statement is not valid, I could not find anything on google regarding the issue, maybe my googling skills just suck.
if mynumber != (0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8) then
print("Please choose an integer number between 1-8")
end
Thanks for any help!!
Correct. That is not how you test things like that. You cannot test multiple values that way.
or requires expressions on either side and evaluates to a single expression. So (0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8) evaluates to 0 and your final expression is just if mynumber != 0 then.
To test multiple values like that you need to use or around multiple comparison expressions.
if (mynumber ~= 0) or (mynumber ~= 1) or (mynumber ~= 2) ... then (also notice ~= is the not-equal operator not !=).
Also be sure to note YuHao's answer about the logic in this line and how to test for this correctly.
Others have pointed the major problems you have, i.e, 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 evaluates as 0, the rest is ignored because of short-circuit. You need to test the number with these numbers one by one.
However, there's one last trap. The condition
if mynumber ~= 0 or mynumber ~= 1 then
is always true, because a number is either not equal to 0, in which case mynumber ~= 0 is true; or it is equal to 0, in which case mynumber ~= 1 is true.
The correct logic should be:
if mynumber ~= 0 and mynumber ~= 1 then
Etan's answer explains the behaviour as observed in lua. I'd suggest writing a custom FindIn function for searching:
function FindIn( tInput, Value )
for _ in pairs( tInput ) do
if Value == tInput[_] then return true end
end
return false
end
if FindIn( {1,2,3,4,5,6,7,8}, mynumber ) then
-- ...
end
try this:
In Lua You check if two items are NOT EQUAL by "~=" instead of "!=",
If You compare two items in if statement, then always remember that items should return booleans, so: instead of mynumber != (0 or 1 or...) try something like (mynumber ~= 0) or (mynumber ~= 1) ...
You can do it simple with .... (mynumber have to be integer variable)
if mynumber<0 or mynumber>8 then
print("Please choose an integer number between 1-8")
end
I am working on generating a 40 bit length pulse train. I also must be able to adjust the frequency. I tried to make a new low frequency clock and i make a new counter which counts on it's rising edges and give an high output and terminating after 40 bit. It's not working. I tried some other methods. They are not, too.
For example;
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.NUMERIC_STD.all;
entity con40 is port(clk:in std_ulogic; q:out std_ulogic);
end entity con40;
architecture Behaviour of con40 is
constant s:std_ulogic_vector:="11111111111111111111111111111111";
signal i:unsigned(4 downto 0):="00000";
signal en:std_logic:='1';
signal reset:std_logic:='0';
begin
q<=s(to_integer(i));
process(reset,clk) is begin
if reset='1' then
i<=(others=>'0');
elsif rising_edge(clk) then
if en='1' then
i<=i+1;
end if;
end if;
end process;
end architecture Behaviour;
There is 32-bit length in this code but i wanna make 40 bit but whatever, this is not working too. I think methods for such a pulse train must be common and they are being used widely. But hey! unluckily i can find nothing useful.
I took the liberty of moving en and reset to port signals, also changed your constant to a recognizable 40 bit value, and specified the range to make it a locally static constant.
The issue with your counter is that it isn't big enough to address 40 bits. You have i specified as a 5 bit value while 40 bits requires a 6 bit counter.
I also added a second architecture here with i as an integer type signal. With i as either an unsigned value or an integer type you likely need to roll over the i counter at 39 ("100111") when the first position is 0 ("000000").
library ieee;
use ieee.std_logic_1164.all;
entity con40 is
port(
reset: in std_ulogic;
clk: in std_ulogic;
en: in std_ulogic;
q: out std_ulogic
);
end entity con40;
architecture foo of con40 is
constant s: std_ulogic_vector (0 to 39) := x"feedfacedb";
signal i: natural range 0 to 39;
begin
q <= s(i);
process (reset, clk)
begin
if reset = '1' then
i <= 0;
elsif rising_edge(clk) and en = '1' then
if i = 39 then
i <= 0;
else
i <= i + 1;
end if;
end if;
end process;
end architecture;
library ieee;
use ieee.numeric_std.all;
architecture behave of con40 is
constant s: std_ulogic_vector (0 to 39) := x"feedfacedb";
signal i: unsigned (5 downto 0);
begin
q <= s(to_integer(i));
process (reset, clk)
begin
if reset = '1' then
i <= "000000";
elsif rising_edge(clk) and en = '1' then
if i = "100111" then
i <= "000000";
else
i <= i + 1;
end if;
end if;
end process;
end architecture;
I also did a quick and dirty test bench:
library ieee;
use ieee.std_logic_1164.all;
entity tb_con40 is
end entity;
architecture foo of tb_con40 is
signal clk: std_ulogic := '0';
signal reset: std_ulogic := '1';
signal en: std_ulogic := '0';
signal q: std_ulogic;
begin
DUT:
entity work.con40
port map (
reset => reset,
clk => clk,
en => en,
q => q
);
CLOCK:
process
begin
for i in 0 to 46 loop
wait for 20 ns;
clk <= not clk;
wait for 20 ns;
clk <= not clk;
end loop;
wait;
end process;
STIMULUS1:
reset <= '0' after 40 ns;
STIMULUS2:
en <= '1' after 60 ns;
end architecture;
Which can demonstrate the correct output:
addendum in response to comment question
The pattern X"FEEDFACEDB" is 40 bits long and was substituted for the 32 all '1's value for constant s to demonstrate that you are actually addressing individual elements of the s array value.
To stop the pulse train fro recurring:
For architecture foo (using an integer type for i):
elsif rising_edge(clk) and en = '1' then
-- if i = 39 then
-- i <= 0;
-- else
if i /= 39 then -- added
i <= i + 1;
end if;
This stops the counter from operating when it reaches 39.
For architecture behave (using an unsigned type for i):
elsif rising_edge(clk) and en = '1' then
-- if i = "100111" then
-- i <= "000000";
-- else
if i /= "100111" then -- added
i <= i + 1;
end if;
end if;
Both architectures behave identically stopping the i counter at 39 ("100111").
The counter can be shown to have stopped by simulating:
Without adding an additional control input the only way to get the pulse stream to occur a second time would be by invoking reseet.
The following code could be a simple implementation to generate pulse trains. This module requires a start impulse (StartSequence) and acknowledges the generated sequence with 'SequenceCompleted'.
Instead of an state machine I use a basic RS flip flop with set = StartSequence and rst = SequenceCompleted_i. I also broke up the process into two processes:
state control - this can be extended to a full FSM if needed
used for counter(s)
Initially, the module emits PULSE_TRAIN(0) by default and also after each sequence generation. So if you want to emit 40 ones otherwise zero set PULSE_TRAIN := (0 => '0', 1 to 40 => '1')
This module is variable in the bit count of PULSE_TRAIN, so I needed to include a function called log2ceil, which calculates the 2s logarithm aka needed bits from PULSE_TRAIN's length attribute.
So in case of 'length = 41 bits Counter_us has a range of (5 downto 0).
entity PulseTrain is
generic (
PULSE_TRAIN : STD_LOGIC_VECTOR
);
port (
Clock : in STD_LOGIC;
StartSequence : in STD_LOGIC;
SequenceCompleted : out STD_LOGIC;
Output : out STD_LOGIC
);
end entity;
architecture rtl of PulseTrain is
function log2ceil(arg : POSITIVE) return NATURAL is
variable tmp : POSITIVE := 1;
variable log : NATURAL := 0;
begin
if arg = 1 then return 0; end if;
while arg > tmp loop
tmp := tmp * 2;
log := log + 1;
end loop;
return log;
end function;
signal State : STD_LOGIC := '0';
signal Counter_us : UNSIGNED(log2ceil(PULSE_TRAIN'length) - 1 downto 0) := (others => '0');
signal SequenceCompleted_i : STD_LOGIC;
begin
process(Clock) is
begin
if rising_edge(Clock) then
if (StartSequence = '1') then
State <= '1';
elsif (SequenceCompleted_i = '1') then
State <= '0';
end if;
end if;
end process;
SequenceCompleted_i <= '1' when (Counter_us = (PULSE_TRAIN'length - 1)) else '0';
SequenceCompleted <= SequenceCompleted_i;
process(Clock)
begin
if rising_edge(Clock) then
if (State = '0') then
Counter_us <= (others => '0');
else
Counter_us <= Counter_us + 1;
end if;
end if;
end process;
Output <= PULSE_TRAIN(to_integer(Counter_us));
end;
As what #fru1tbat mentioned, it's not really clear what is "not working" and what you really intend to do. If you would really just want to generate a pulse train, one would think you want to generate a series of alternating '1' and '0', not all '1's like in the code you posted.
Also, the i counter just counts up, and can only be reset to '0' by use of the reset signal, which is fine as long as you intended it that way.
If you'd like to generate a train of '1's and '0's, you'd need something like this (not tested, but should be along these lines):
architecture behaviour of con40 is
constant trainLength:positive:=80;
signal i:unsigned(6 downto 0):=(others=>'0');
...
begin
process(reset,clk) is begin
if reset then
i<=(others=>'0');
q<='0';
elsif rising_edge(clk) then
q<='0'; -- default assignment.
-- Defaults to '0' when if-statement fails.
if i<trainLength then
i<=i+1;
q<=not q;
end if;
end if;
end process;
end architecture behaviour;
This gives you a single-shot pulse train, means there is no way to repeat generation of the pulse train unless you assert the reset signal again. This is fine if it's what you want, otherwise, you'll need more signals to cater for cases where you'd like to re-generate the pulse train without resetting.
Here, I'm assuming you'd like 40 HIGH pulses, which essentially makes the train length 80 clock cycles, not 40. Also, I'm assuming you want a 50% duty cycle, i.e. the HIGH and LOW times are equal. Depending on your requirements, you may need a pulse width that is longer or shorter.
With these assumptions in mind, you'd need at least a 7-bit counter to count 80 clocks. You may think of other better ways to do this as well, but this just comes off the top of my head, and is probably a good place to start.
If your tool doesn't yet support VHDL-2008's enhanced port modes (e.g. ability to read from out-mode ports), then you could declare q as having a buffer mode instead of out. If your tool doesn't support buffer port modes, then you can declare an internal signal and use it for your logic. E.g.:
signal i_q: std_ulogic;
...
i_q<=not i_q; -- use internal signal for logic instead.
q<=i_q; -- drive output from internal signal.
To adjust the frequency, simply supply a higher or lower frequency into your clk input. This can be generated from another PLL, or a frequency divider, or any other oscillating circuitry you have available. Just supply its output into your clk.
Hope this helps.