ISE 自动推断 RAM 块需要满足哪些要求？

Question

我有一段 IP，它应该是 32 位字节可寻址内存。但我无法让它推断块内存，它正在推断大量的触发器...

它应该适合仅具有双端口块内存的 Spartan3e (xc3s1200e-4fg320)，实际上内存被分为两个数组按奇偶排列...

这是代码，我希望可以帮助理解我做错了什么？

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;

package mem_types is
    type memory_t is array (natural range <>) of std_logic_vector(7 downto 0);
end mem_types;

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
use work.mem_types.all;

---- Uncomment the following library declaration if instantiating
---- any Xilinx primitives in this code.
--library UNISIM;
--use UNISIM.VComponents.all;

entity ram is
    generic (
        INIT : memory_t(0 to 4095) := (others => (others => '0'))
    );

    port ( clk, rst : in std_logic;
             addr : in std_logic_vector(11 downto 0);
             din : in std_logic_vector(31 downto 0);
             dout : out std_logic_vector(31 downto 0);
             we : std_logic_vector(3 downto 0)
          );
end ram;

architecture Behavioral of ram is
    type ramport_t is record
        addr : std_logic_vector(10 downto 0);
        dout : std_logic_vector(7 downto 0);
        din : std_logic_vector(7 downto 0);
        wea : std_logic;
    end record;
    signal port0a, port0b, port1a, port1b : ramport_t;
    signal addr_a, addr_b, addr_c, addr_d : std_logic_vector(11 downto 0);
    signal memory0, memory1 : memory_t(0 to 2047);
begin

    addr_a <= addr;
    addr_b <= addr+1;
    addr_c <= addr+2;
    addr_d <= addr+3;

    port0a.addr <= addr_a(11 downto 1) when addr_a(0) = '0' else addr_b(11 downto 1);
    port1a.addr <= addr_b(11 downto 1) when addr_b(0) = '1' else addr_a(11 downto 1);
    port0b.addr <= addr_c(11 downto 1) when addr_c(0) = '0' else addr_d(11 downto 1);
    port1b.addr <= addr_d(11 downto 1) when addr_d(0) = '1' else addr_c(11 downto 1);

    dout(07 downto 00) <= port0a.dout when addr_a(0) = '0' else port1a.dout;
    dout(15 downto 08) <= port1a.dout when addr_b(0) = '1' else port0a.dout;
    dout(23 downto 16) <= port0b.dout when addr_c(0) = '0' else port1b.dout;
    dout(31 downto 24) <= port1b.dout when addr_d(0) = '1' else port0b.dout;

    port0a.din <= din(07 downto 00) when addr_a(0) = '0' else din(15 downto 08);
    port1a.din <= din(15 downto 08) when addr_b(0) = '1' else din(07 downto 00);
    port0b.din <= din(23 downto 16) when addr_c(0) = '0' else din(31 downto 24);
    port1b.din <= din(31 downto 24) when addr_d(0) = '1' else din(23 downto 16);

    port0a.wea <= we(0) when addr_a(0) = '0' else we(1);
    port1a.wea <= we(1) when addr_b(0) = '1' else we(0);
    port0b.wea <= we(2) when addr_c(0) = '0' else we(3);
    port1b.wea <= we(3) when addr_d(0) = '1' else we(2);

    port0a.dout <= memory0(conv_integer(port0a.addr));
    port0b.dout <= memory0(conv_integer(port0b.addr));
    port1a.dout <= memory1(conv_integer(port1a.addr));
    port1b.dout <= memory1(conv_integer(port1b.addr));

    process (clk, rst)
    begin
        if rst = '1' then
            for a in 0 to 2047 loop
                memory0(a) <= INIT(a*2);
            end loop;
        elsif falling_edge(clk) then
            if (port0a.wea = '1') then
                memory0(conv_integer(port0a.addr)) <= port0a.din;
            end if;

            if (port0b.wea = '1') then
                memory0(conv_integer(port0b.addr)) <= port0b.din;
            end if;
        end if;
    end process;

    process (clk, rst)
    begin
        if rst = '1' then
            for a in 0 to 2047 loop
                memory1(a) <= INIT((a*2)+1);
            end loop;
        elsif falling_edge(clk) then
            if (port1a.wea = '1') then
                memory1(conv_integer(port1a.addr)) <= port1a.din;
            end if;

            if (port1b.wea = '1') then
                memory1(conv_integer(port1b.addr)) <= port1b.din;
            end if;
        end if;
    end process;

end Behavioral;

Answer 1

Xilinx 综合指南 下的 FPGA 流程编码。我几乎可以肯定重置循环会导致推断失败。该代码需要同时访问存储器的所有元素，而这对于 Block RAM 来说是不可能的。

Answer 2

你不能这样做：

process (clk, rst)
begin
    if rst = '1' then
        for a in 0 to 2047 loop
            memory0(a) <= INIT(a*2);
        end loop;

...因为这需要一个可重置内存，而不是一个初始化内存。

要初始化，您需要将信号声明更改为以下形式

signal memory0 : memory_t(0 to 2047) := ( some list of integers or something that returns an array of integers);

您当前执行此操作的方式（与您的 init 交错）意味着您必须使用一个函数：（

function init_mem(init_values: memory_t) returns memory_t is
variable retval : memory_t(init_values'high/2)+1 downto 0);
begin
  for i in retval'range loop
      retval(i) := init_values(2*i);
  end for;
end function;

请注意，这是我凭空想象出来的我什至没有尝试编译它，所以对任何拼写错误和语法错误表示歉意......但我希望你明白:)

然后你可以用它来初始化信号：

signal memory0 : memory_t(0 to 2047) := init_mem(INIT);

这将全部用于模拟。您可能会或可能不会成功使用 XST 合成器推断 INIT 值 - 我还没有尝试过。检查综合日志文件以查看其报告的内容 - 请向我们报告它是否有效以及您在哪个版本的 XST 上尝试过它。