两个二进制数的除法，每个一个是4位，并输出浮点

Question

此代码使两个二进制数的除法运算每个数字为4位，我需要修改此代码以使输出分数结果为例，例如，说明我的想法1010（10）除以0011（3）= 11.010101（3.3333） ），那么需要修改什么部分才能在电路输出中以二进制形式显示结果 预先谢谢你

 library IEEE;
 use IEEE.STD_LOGIC_1164.ALL;
 use IEEE.STD_LOGIC_unsigned.All;

 entity division4bits is

 generic(SIZE: INTEGER := 4); 
 port(reset: in STD_LOGIC; 
    en: in STD_LOGIC; 
    clk: in STD_LOGIC; 

    num: in STD_LOGIC_VECTOR((SIZE - 1) downto 0); 
    den: in STD_LOGIC_VECTOR((SIZE - 1) downto 0); 
    result: out STD_LOGIC_VECTOR((SIZE - 1) downto 0); 
    remender: out STD_LOGIC_VECTOR((SIZE - 1) downto 0) 
    );

  end division4bits;




 architecture Behavioral of division4bits is

signal buf: STD_LOGIC_VECTOR((2 * SIZE - 1) downto 0); 
signal dbuf: STD_LOGIC_VECTOR((SIZE - 1) downto 0); 
signal sm: INTEGER range 0 to SIZE; 

alias buf1 is buf((2 * SIZE - 1) downto SIZE); 
alias buf2 is buf((SIZE - 1) downto 0); 
begin 
p_001: process(reset, en, clk) 
begin 
    if reset = '1' then 
        result <= (others => '0'); 
        remender <= (others => '0'); 
        sm <= 0; 
       elsif rising_edge(clk) then 
        if en = '1' then 
            case sm is 
            when 0 => 
                buf1 <= (others => '0'); 
                buf2 <= num; 
                dbuf <= den; 
                result <= buf2; 
                remender <= buf1; 
                sm <= sm + 1; 
              when others => 
                if buf((2 * SIZE - 2) downto (SIZE - 1)) >= dbuf then 
                  buf1 <= '0' & (buf((2 * SIZE - 3) downto (SIZE - 1)) - 
      dbuf((SIZE - 2) downto 0)); 
                    buf2 <= buf2((SIZE - 2) downto 0) & '1'; 
                else 
                    buf <= buf((2 * SIZE - 2) downto 0) & '0'; 
                end if; 
                if sm /= SIZE then 
                    sm <= sm + 1; 
                else 
                    sm <= 0; 
                end if; 
            end case; 
        end if; 
    end if; 
    end process; 


     end Behavioral;