VHDL n 位桶形移位器

例如 shift = 5d -> 00101b 使第 1 阶段中的多路复用器能够移位 1 位，使第 3 阶段中的多路复用器能够移位 4 位.所有其他多路复用阶段都设置为通过(shift(i) = 0).我也不建议将基本换档与换档模式(算术、逻辑、旋转)和方向(左、右)混为一谈.算术和逻辑的区别只是移入值右移可以通过转换完成 => shiftright = reverse(shiftleft(reverse(input), n)可以在此处找到开源实现:https://github.com/VLSI-EDA/PoC/blob/master/src/arith/arith_shifter_barrel.vhdlI have a 32 bit barrel shifter using behavior architecture. Now I need to convert it to an n-bit shifter. The problem that I'm facing is that there is some kind of restriction to the for loop that I have to put a constant as sentinel value.Following is my Codelibrary IEEE;use IEEE.std_logic_1164.all;Entity bshift is -- barrel shifter port (left : in std_logic; -- '1' for left, '0' for right logical : in std_logic; -- '1' for logical, '0' for arithmetic shift : in std_logic_vector(4 downto 0); -- shift count input : in std_logic_vector (31 downto 0); output : out std_logic_vector (31 downto 0) );end entity bshift;architecture behavior of bshift is function to_integer(sig : std_logic_vector) return integer is variable num : integer := 0; -- descending sig as integer begin for i in sig'range loop if sig(i)='1' then num := num*2+1; else num := num*2; end if; end loop; -- i return num; end function to_integer;begin -- behavior shft32: process(left, logical, input, shift) variable shft : integer; variable out_right_arithmetic : std_logic_vector(31 downto 0); variable out_right_logical : std_logic_vector(31 downto 0); variable out_left_logical : std_logic_vector(31 downto 0); begin shft := to_integer(shift); if logical = '0' then out_right_arithmetic := (31 downto 32-shft => input(31)) & input(31 downto shft); output <= out_right_arithmetic after 250 ps; else if left = '1' then out_left_logical := input(31-shft downto 0) & (shft-1 downto 0 => '0'); output <= out_left_logical after 250 ps; else out_right_logical := (31 downto 32-shft => '0') & input(31 downto shft); output <= out_right_logical after 250 ps; end if; end if; end process shft32;end architecture behavior; -- of bshiftany help will be appreciated 解决方案 Your code is not a barrel shifter implementation, because a barrel shift is a mux-tree.If you have a 32 bit BarrelShifter module, you will need a 5 bit Shift input, wherein every bit position i enables a 2^i shift operation.So for example shift = 5d -> 00101b enables a mux in stage 1 to shift for 1 bit and a mux in stage 3 to shift 4 bits. All other mux stages are set to pass through (shift(i) = 0).I also would not advice to mix up basic shifting with shift modes (arithmetic, logic, rotate) and directions (left, right).arithmetic and logic is only different in the shift-in valueshift right can be done by a conversion => shiftright = reverse(shiftleft(reverse(input), n)An open source implementation can be found here:https://github.com/VLSI-EDA/PoC/blob/master/src/arith/arith_shifter_barrel.vhdl 这篇关于VHDL n 位桶形移位器的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！