我有一个带有
D_IN_VAL std_logic_vector(n-1 downto 0)
输入端口和内部 signal CUR_MAX_S : std_logic_vector(n-1 downto 0)
的模块。我希望仅使用部分 D_IN_VAL
数据(例如 20 位信号中的 4 位),因此我想将 D_IN_VAL
“保存”在 CUR_MAX_S
中,然后使用部分 CUR_MAX_S
。我已经将 D_IN_VAL
同步分配给 CUR_MAX_S
if RISING_EDGE(CLK) then
并且在某些条件下 if D_SEND = '1' AND WHAT_STATE = 0 then
。
但是当
D_IN_VAL
的值发生变化时会发生什么? CUR_MAX_S
的值也会改变吗?或者 Quartus 是否生成一个锁存器,它会记住我分配给 CUR_MAX_S 的内容?根据模拟,它“记住”了该值,但我已经学会了不要 100% 相信模拟。它是一个信号而不是变量,所以我不确定。
编辑:添加示例代码,更改相关变量的名称以匹配代码
library IEEE;
use IEEE.STD_LOGIC_1164.all;
use ieee.numeric_std.all;
entity MYMUX1 is
generic (
-- dlugosc rejestru jako potega 2-ki, nie mniej niz 3
-- 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32, 2^6 = 64, 2^7 = 128, 2^8 = 256,
M : INTEGER := 4;
-- szerokosc we/wy rejestru (w bitach)
N : INTEGER := 5
);
port(
-- wartosc maksimum
D_IN_VAL : in STD_LOGIC_VECTOR(2*N+M-1 downto 0);
-- pozycja maksimum
D_IN_POS : in STD_LOGIC_VECTOR(M+1 downto 0);
-- wyslij nastepne dane
D_SEND : in STD_LOGIC;
-- zegar, duh.
CLK : in STD_LOGIC;
-- resET
RST : in STD_LOGIC;
-- dane wyjsciowe, do UARTu
D_OUT : out STD_LOGIC_VECTOR(7 downto 0);
-- dane gotowe
D_RDY : out STD_LOGIC;
-- testowe
test1 : out STD_LOGIC_VECTOR(11 downto 0);
test2 : out STD_LOGIC_VECTOR(11 downto 0);
test3 : out STD_LOGIC_VECTOR(11 downto 0);
test4 : out STD_LOGIC_VECTOR(11 downto 0);
test5 : out STD_LOGIC_VECTOR(3 downto 0);
test_state : out STD_LOGIC_VECTOR(3 downto 0);
test_neg : out STD_LOGIC_VECTOR(2*N+M-1 downto 0)
);
end MYMUX1;
architecture AMYMUX1 of MYMUX1 is
signal CUR_MAX_S : STD_LOGIC_VECTOR(2*N+M-1+4 downto 0);
signal WHERE_MAX_S : STD_LOGIC_VECTOR(M+1+4 downto 0);
begin
process(CLK)
-- ile bajtow na wartosc? maks 4mln, 2^32. na przyklad n=12 bajtow, 2^m, m=8, 256 razy
variable COUNT1 : INTEGER range 0 to 7;
-- ile bajtow na pozycje? maks 65k, 2^16
variable COUNT2 : INTEGER range 0 to 3;
variable COUNT_MAIN : INTEGER range 0 to 11;
variable CUR_MAX : INTEGER range -(2**(2*N+M)) to (2**(2*N+M)-1);
variable WHERE_MAX : INTEGER range 0 to (2*2**M-1);
variable WHAT_STATE : INTEGER range 0 to 6; -- albo 6
begin
-- if(RISING_EDGE(CLK)) then
if(((2*N+M) mod 4)>0) then
COUNT1 := ((2*N+M)/4)+1;
else
COUNT1 := (2*N+M)/4;
end if;
if(((M+1+1) mod 4)>0) then
COUNT2 := ((M+1+1)/4)+1;
else
COUNT2 := (M+1+1)/4;
end if;
if(RISING_EDGE(CLK)) then
if(RST = '1') then
WHAT_STATE := 0;
test_state <= STD_LOGIC_VECTOR(to_signed(0,4));
else
-- state 0 - jesli wartosc > od zera, zapisz, jesli mniejsza, odwroc i zapisz, wyslij ! jako znak poczatku paczki
if(D_SEND = '1' AND WHAT_STATE = 0) then
if(to_integer(signed(D_IN_VAL))>=0) then
-- CUR_MAX := to_integer(signed(D_IN_VAL));
CUR_MAX_S <= "0000" & STD_LOGIC_VECTOR((signed(D_IN_VAL)));
-- skocz od razu do wysylania danych
WHAT_STATE := 2;
else
-- CUR_MAX := -to_integer(signed(D_IN_VAL));
CUR_MAX_S <= "0000" & STD_LOGIC_VECTOR((-signed(D_IN_VAL)));
-- wyslij znak a dopiero potem dane
WHAT_STATE := 1;
end if;
WHERE_MAX_S <= "0000" & D_IN_POS;
test_neg <= STD_LOGIC_VECTOR(-signed(D_IN_VAL));
WHERE_MAX := to_integer(signed(D_IN_POS));
COUNT_MAIN := COUNT1;
D_OUT <= STD_LOGIC_VECTOR(to_unsigned(33,8));
test1 <= STD_LOGIC_VECTOR(to_signed(CUR_MAX,12));
test2 <= STD_LOGIC_VECTOR(to_signed(WHERE_MAX,12));
test3 <= STD_LOGIC_VECTOR(to_signed(COUNT_MAIN,12));
test4 <= STD_LOGIC_VECTOR(to_signed(WHAT_STATE,12));
test_state <= STD_LOGIC_VECTOR(to_signed(0,4));
-- oba teoretycznie dzialaja, ale ktore jest poprawne...?
-- CUR_MAX_S <= STD_LOGIC_VECTOR(to_unsigned(CUR_MAX,14));
-- CUR_MAX_S <= D_IN_VAL;
-- state 1 - wyslij minus, jesli liczba jest ujemna
elsif(WHAT_STATE = 1) then
-- and so on, and so forth... the rest of the code goes here.
根据您的描述而不是您的长代码:
CUR_MAX_S
将被合成为 n 位寄存器,其中 CLK
信号路由到组合逻辑的时钟输入和输出 D_SEND = '1' AND WHAT_STATE = 0
路由到每个位的使能输入。