SAP computer之RAM
RAM
The RAM is a 16 X 8 static TTL RAM. We can program the RAM by means of the address and data switch registers. This allows us to store a program and data in the memory before a computer run.
During a computer run, the RAM receive 4-bit addresses from MAR and a read operation is performed. In this way, the instuction or data word stored in the RAM is placed on the W bus for use in some other part of the computer.
library IEEE;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all; entity ROM_16_8 is
port
(
READ : in std_logic; --! Active low enable ROM signal, (tri-state)
ADDRESS : in std_logic_vector ( downto ); --! -bit ROM address bits from MAR
DATA_OUT : out std_logic_vector ( downto ) --! -bit ROM output word to W-bus
);
end ROM_16_8 ; architecture beh of ROM_16_8 is type mem is array ( to ) of std_logic_vector( downto ) ;
signal rom : mem; begin
20 --! This program works as follow:
21 --!
22 --! Load 5 to AC (memory content of 9)
23 --! Output 5 (content of AC)
24 --! Add 7 (memory content of 10) to 5 (AC content)
25 --! Output 12 (content of AC)
26 --! Add 3 (memory content of 11) to 12 (AC content)
27 --! Subtract 4 (memory content of 12) from 15 (AC content)
28 --! Output 11 (content of AC)
rom <= (
=> "" , -- LDA 9h ... Load AC with the content of memory location 9
=> "" , -- OUT
=> "" , -- ADD Ah ... Add the contents of memory location A to the AC content and replace the AC
=> "" , -- OUT
=> "" , -- ADD Bh ... Add the contents of memory location B to the AC content and replace the AC
=> "" , -- SUB Ch ... Sub the contents of memory location C from the AC content and replace the AC
=> "" , -- OUT
=> "" , -- HLT
=> "" ,
=> "" , --5
=> "" , --7
=> "" , --3
=> "" , --4
=> "" ,
=> "" ,
=> "" ); process (READ,ADDRESS)
begin
if READ = '' then
DATA_OUT <= rom(to_integer(unsigned(ADDRESS))) ;
else
DATA_OUT <= (DATA_OUT'range => 'Z');
end if;
end process ; end beh;
SAP computer之RAM的更多相关文章
- SAP computer之input and MAR
Input and MAR Below the program counter is the input and MAR block. It includes the address and data ...
- SAP computer之program counter
Program counter The program is stored in memory with the first instruction at binary address 0000, t ...
- SAP computer之architecture
Simple-As-Possible computer introduces all the cruicial ideas behind computer operation without bury ...
- VMware12 安装 CentOS 6.5 64位
前言:本人在配置Hadoop的过程中,需要搭建Cent OS 64 环境,借此,顺便将Cent OS 64 的安装在此记录,方便自己,也方便大家学习.本次是在VM12虚拟机中实现Cent OS 64 ...
- 堆栈 & Stack and Heap
What's the difference between a stack and a heap? The differences between the stack and the heap can ...
- PostgreSQL Hardware Performance Tuning
Bruce Momjian POSTGRESQL is an object-relational database developed on the Internet by a group of de ...
- MongoDB十二种最有效的模式设计【转】
持续关注MongoDB博客(https://www.mongodb.com/blog)的同学一定会留意到,技术大牛Daniel Coupal 和 Ken W. Alger ,从 今年 2月17 号开始 ...
- Client Dataset Basics
文章出处: http://www.informit.com/articles/article.aspx?p=24094 In the preceding two chapters, I discus ...
- What’s the difference between a stack and a heap?
http://www.programmerinterview.com/index.php/data-structures/difference-between-stack-and-heap/ The ...
随机推荐
- 2.2 为什么要使用Shell脚本
使用脚本编程语言的好处是,它们多半运行在比编译型语言还高的层级,能够轻易处理文件与目录之类的对象.缺点是:它们的效率通常不如编译型语言.不过权衡之下,通常使用脚本编程还是值得的:花一个小时写成 ...
- 平衡树前置——BST
上一节:平衡树——序 BST(Binary Search Tree)二叉排序树,其定义为:二叉排序树或者是空树,或者是满足如下性质的二叉树: ①若它的左子树非空,则左子树上所有结点的值均小于根结点的值 ...
- 【DIP Learining MFC &OpenCV】 Experience by 20171026
This day saw the progress I achieved in creating a fusion of MFC frame and OpenCV code as well as so ...
- SpringMVC demo 小例子,实现简单的登录和注册
1.创建一个动态的web工程 2.导入springMvc所需要的jar包(这里可以去网上找,资源有很多) 前两部就不详细描述了,后面才是正经代码~ 首先有一个web.xml文件,这个属于大配置文件,由 ...
- 2.1.6、SparkEnv中创建ShuffleManager
ShuffleManager负责管理本地以及远程的block数据的shuffle操作. ShffuleManager的创建是在SparkEnv中. // Let the user specify sh ...
- linux -- 视频尺寸-cif、2cif、dcif、D1、HD1、4D1
1 CIF简介 CIF是常用的标准化图像格式(Common Intermediate Format).在H.323协议簇中,规定了视频采集设备的标准采集分辨率.CIF = 352×288像素 ...
- [bzoj4003][JLOI2015]城池攻占_左偏树
城池攻占 bzoj-4003 JLOI-2015 题目大意:一颗n个节点的有根数,m个有初始战斗力的骑士都站在节点上.每一个节点有一个standard,如果这个骑士的战斗力超过了这个门槛,他就会根据城 ...
- HDU 4524
简单题,先从右边消起,注意结束时a[1]==0才能是yes #include <iostream> #include <cstdio> #include <cstring ...
- POJ2584_T-Shirt Gumbo(二分图多重最大匹配/最大流)
解题报告 http://blog.csdn.net/juncoder/article/details/38239367 题目传送门 题意: X个參赛选手,每一个选手有衣服大小的范围,5种大小的队服,求 ...
- 计算几何 二维凸包问题 Andrew算法
凸包:把给定点包围在内部的.面积最小的凸多边形. Andrew算法是Graham算法的变种,速度更快稳定性也更好. 首先把全部点排序.依照第一keywordx第二keywordy从小到大排序,删除反复 ...