我正在尝试在VHDL中创建5维数组,但不确定如何设置和初始化位。
这是我到目前为止所拥有的:
type \1-line\ is array (4 - 1 downto 0) of unsigned (32 - 1 downto 0);
type square is array (4 - 1 downto 0) of \1-line\;
type cube is array (4 - 1 downto 0) of square;
type hypercube is array (4 - 1 downto 0) of cube;
type \5-cube\ is array (4 - 1 downto 0) of cube;
signal mega_array : \5-cube\;
begin
process (clock, reset) begin
if (reset == '1') then
mega_array <= '0';
end if;
end process;
end behv;
最佳答案
一种实现方法是使用'(others =>'0')'。这是一种将向量的所有位设置为0的干净安全的方法。您必须对阵列的每一层都执行此操作。
library ieee;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all;
entity test is
port (
clock : in std_logic;
reset : in std_logic);
end entity test;
architecture behv of test is
type \1-line\ is array (4 - 1 downto 0) of unsigned (32 - 1 downto 0);
type square is array (4 - 1 downto 0) of \1-line\;
type cube is array (4 - 1 downto 0) of square;
type \5-cube\ is array (4 - 1 downto 0) of cube;
signal mega_array : \5-cube\;
begin
process (clock, reset)
begin
if (reset = '1') then -- note: not '=='
mega_array <= (others => (others => (others => (others => (others => '0')))));
end if;
end process;
end architecture behv;
请注意,尽管
\1-...
命名是正确的VHDL,但我不会使用它来避免讨厌的工具问题。我不确定它们是否会来,但是避免它们比解决它们更好。我会改用t_1line
。关于arrays - 如何:VHDL中的多维数组,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/14894204/