问题描述
我正在尝试在VHDL中创建一个32位计数器。下面是我的代码：
I'm trying to make a 32 bit counter in VHDL. Below is my code:
LIBRARY ieee;
USE ieee.std_logic_1164.ALL;
USE ieee.numeric_std.ALL;
ENTITY counter32 IS
PORT (en, clk, clr: IN STD_LOGIC;
count: OUT STD_LOGIC_VECTOR(4 DOWNTO 0));
END counter32;
ARCHITECTURE rtl OF counter32 IS
SIGNAL count_result: STD_LOGIC_VECTOR(4 DOWNTO 0);
BEGIN
counter32: PROCESS(clk, clr)
BEGIN
count <= "00000"; --Initialize counter to all zeroes
IF (clr = '0') THEN
count_result <= "00000";
ELSIF (clk = '1' and clk'EVENT) THEN
IF (en = '1') THEN
count <= STD_LOGIC_VECTOR(unsigned(count_result) + 1);
count <= STD_LOGIC_VECTOR(count_result);
ELSIF (count_result = "11111") THEN
count_result <= "00000";
END IF;
END IF;
END PROCESS counter32;
END rtl;
我的测试台代码在这里：
My test bench code is here:
library ieee;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all;
entity counter32_tb is
end counter32_tb;
architecture io of counter32_tb is
component counter32 is
port(en,clk,clr:in std_logic; count:out std_logic_vector(4 downto 0));
end component;
for all: counter32 use entity work.counter32(rtl);
signal en,clk,clr:std_logic;
signal count:std_logic_vector(4 downto 0);
begin
count <= "00000";
g0: counter32 port map(en,clk,clr,count);
p0: process
begin
en <= '1';
clk <= '0';
clr <= '1';
wait for 10ns;
en <= '1';
clk <= '1';
clr <= '1';
wait for 10ns;
en <= '1';
clk <= '0';
clr <= '1';
wait for 10ns;
en <= '1';
clk <= '1';
clr <= '1';
wait for 10ns;
en <= '1';
clk <= '0';
clr <= '1';
wait for 10ns;
en <= '1';
clk <= '1';
clr <= '0';
end process;
end io;
但是，每当我进行测试时，加1都会得到'U'STD_LOGIC值和红色条形在测试中，如您在这里看到的：
Whenever I test, however, an addition of 1 gives a 'U' STD_LOGIC value and a red bar in testing, as you can see here:
你知道这是怎么回事吗？我真的很困惑！
Any idea what the matter is? I'm really confused!
推荐答案
你知道这是怎么回事吗？
您的波形与您的测试台激励不匹配。
Any idea what the matter is?
Your waveform doesn't match your test bench stimulus.
信号计数共有三个分配，它们似乎显示在您的波形中（在测试台上） ）。初始分配为 00000，以及两个条件分配。来回跳动是由于进程对clk敏感，导致使用第一条赋值语句在时钟的下一个边缘跳回 00000。
There are three assignments to the signal count which appears to show in your waveform (at the test bench level). An initial assignment to "00000", and two conditional assignments. The bouncing back and forth is caused by the process sensitity to clk, bouncing back to "00000" on the following edge of clock using the first assignment statement.
声明最后一个分配是生效的分配。您将其写入 00000，然后根据时钟的上升沿有条件地将其更改为count_result。请注意，您实际上也没有以count_result + 1加载count，下一个分配将提供count_result的当前值。虽然我们也不需要将类型转换为std_logic_vector，但count_result已经是一个std_logic_vector。
In a process statement the last assignment is the one that takes effect. You're writing it to "00000" and changing that to count_result conditionally based on the positive edge of clock. Note that you aren't actually loading count with count_result + 1 either, the next assignment provides the current value of count_result. While we're on the subject the type conversion to std_logic_vector isn't needed either, count_result is already a std_logic_vector.
时钟上未知的（红色）闪光优势是因为您实际上尚未清除count_result。 clr上唯一的事件是从'U'到'1'，并且不会清除。
The unknown (red) 'flash' at the clock edge is because you haven't actually cleared count_result. The only event on clr is from 'U' to '1' and causes no clear.
vhdl设计代码无法用作计数器。
The vhdl design code is not functional as a counter.
此：
library ieee;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all;
entity counter32 is
port (
en, clk, clr: in std_logic;
count: out std_logic_vector(4 downto 0)
);
end counter32;
architecture rtl of counter32 is
signal count_result: std_logic_vector(4 downto 0);
begin
counter: process(clk, clr)
begin
if clr = '0' then
count_result <= (others => '0');
elsif clk = '1' and clk'event and en = '1' then
count_result <= std_logic_vector(unsigned(count_result) + 1);
end if;
end process;
count <= count_result;
end rtl;
library ieee;
use ieee.std_logic_1164.all;
entity counter32_tb is
end entity;
architecture foo of counter32_tb is
signal en: std_logic:= '0';
signal clr: std_logic:= '1';
signal clk: std_logic:= '0';
signal count: std_logic_vector (4 downto 0);
begin
DUT: entity work.counter32
port map (
en => en,
clk => clk,
clr => clr,
count => count
);
CLOCK:
process
begin
wait for 10 ns;
clk <= not clk;
if Now > 720 ns then
wait;
end if;
end process;
STIMULUS:
process
begin
clr <= '0';
en <= '1';
wait for 20 ns;
clr <= '1';
wait for 20 ns;
wait for 20 ns;
wait for 20 ns;
wait for 20 ns;
wait for 20 ns;
en <= '0';
wait for 20 ns;
en <= '1';
wait;
end process;
end architecture;
为此：
计数中不需要结束符的原因算术是由于无符号 +运算符的工作原理而引起的，因此在程序包主体中为numeric_std调用了unsigned_add。结束计数是您需要担心的标量增量。
The reason no end cases are necessary in the count arithmetic are due to how the unsigned "+" operator works, calling unsigned_add in the package body for numeric_std. End counts are something you need to worry about in scalar increments.
进行计数（退出）和 count_result的目的是允许在IEEE Std 1076-2008之前的VHDL版本内部读取的计数值。对于符合-2008的模拟，您应该仅使用count。请注意，上面显示的模拟将在早期版本的VHDL上运行。
The purpose behind having count (mode out) and count_result, is to allow the count value to be read internally for versions of VHDL predating IEEE Std 1076-2008. For a -2008 compliant simulation you should be only using count. Note the above simulation shown will run on earlier versions of VHDL.
您也可以将count_result设置为变量。
You could likewise make count_result a variable.
我相信您根据信号阵列的大小知道这是一个5位计数器而不是32位计数器。转换为后者相对容易。
And I trust you're aware based on signal array sizes this is a 5 bit counter and not a 32 bit counter. Converting to the latter is relatively easy.
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