jemas963852741 发表于 2009-11-28 17:23:08

菜鸟~~ 怎么消除竞争冒险的毛刺 半加器而已

library ieee;
use ieee.std_logic_1164.all;

entity adder2 is
port(num1,num2:in std_logic;
       res,carry:out std_logic);
end adder2;

architecture behave of adder2 is
begin
res <= num1 xor num2;
carry <= num1 and num2;
end behave;

http://cache.amobbs.com/bbs_upload782111/files_22/ourdev_508505.jpg
时序仿真 (原文件名:12.jpg)

ngzhang 发表于 2009-11-28 23:09:35

不用管。
页: [1]
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