我为所显示的状态图编写了一个VHDL代码(因为我是新用户,因此无法发布图像)。但是,当我编译它时,它说有错误: 第16行:进程(clk) - 解析时检测到语法错误 第21行:else - 在解析时检测到语法错误 第23行:结束如果; - 解析时检测到语法错误。
这是我的代码:
library IEEE;
use IEEE.std_logic_1164.all;
use IEEE.std_logic_arith.all;
use IEEE.std_logic_unsigned.ALL;
entity memory_controller is
port(clk: in std_logic;
reset: in std_logic;
bus_id: in std_logic_vector(7 downto 0);
read_write, burst: in std_logic;
oe, we, addr_1, addr_2: out std_logic
);
end memory_controller;
architecture behavioral of memory_controller is
type statetype is (idle, decision, wr, rd1, rd2, rd3, rd4);
signal present_state, next_state : statetype;
process(clk) [LINE 16]
begin
if (rising_edge(clk)) then
if (reset ='0') then
present_state <= next_state;
else [LINE 21]
present_state <= idle;
end if; [LINE 23]
end if;
end process;
process(present_state, read_write, ready, burst)
begin
case present_state is
when idle =>
oe => '0'; we=> '0'; addr_1=> '0'; addr_2=> '0';
if(bus_id = "11110011") then
next_state <= decision;
else
next_state <= idle;
end if;
when decision =>
if (read_write = '1')
then next_state <= rd1;
else next_state <= wr;
end if;
when wr =>
we = '1';
if (ready = '1')
then next_state <= idle;
else
next_state <= wr;
end if;
when rd1 =>
oe = '1';
addr_1 = addr_1 + '1';
addr_2 = addr_2 + '1';
if(ready = '0') then
next_state <= rd1;
if(burst = '0') then
next_state <= idle;
else next_state <= rd2;
end if;
when rd2 =>
oe = '1';
addr_1 = addr_1 + '1';
addr_2 = addr_2 + '1';
if(ready = '1') then
next_state => rd3;
else
next_state => rd2;
end if;
when rd3 =>
oe = '1';
addr_1 = addr_1 + '1';
addr_2 = addr_2 + '1';
if(ready = '1') then
next_state => rd4;
else
next_state => rd3;
when rd4 =>
oe = '1';
addr_1 = addr_1 + '1';
addr_2 = addr_2 + '1';
if(ready = '1')
then next_state => idle;
else next_state => rd4;
end if;
end case;
end process;
end behavioral;
语法完全正确,我不明白为什么这是一个错误。可能有什么不对?
另外,我想在ready = 0,burst = 0和ready = 0以及burst = 1的情况下使用assert语句,但是我不太确定如何在主代码中实现它们。
我突出显示第16,21和23行。
任何帮助都会很棒。
答案 0 :(得分:4)
VHDL模块的形式通常是:
entity MODULENAME is
<Port description>
end MODULENAME;
architecture behavioral of MODULENAME is
<signal declarations and similar>
begin
<synchronous and combinatorial logic statements>
end architecture behavioral;
您缺少的是信号声明后的begin
。也就是说,改变
architecture behavioral of memory_controller is
type statetype is (idle, decision, wr, rd1, rd2, rd3, rd4);
signal present_state, next_state : statetype;
process(clk) [LINE 16]
begin
if (rising_edge(clk)) then
要:
architecture behavioral of memory_controller is
type statetype is (idle, decision, wr, rd1, rd2, rd3, rd4);
signal present_state, next_state : statetype;
begin
process(clk) [LINE 16]
begin
if (rising_edge(clk)) then
答案 1 :(得分:3)
正如Sonicwave指出的那样,您在第一次发言之前就错过了begin
关键字。
还有几个语法错误。 信号分配使用左箭头:
oe => '0';
而是oe <= '0';
we = '1';
而是we <= '1';
如果您使用带有直接编译器反馈的编辑器,您将节省大量时间。