INSTRUCTION CYCLE
COMPUTER ORGANIZATION AND ARCHITECTURE DESIGNING FOR PERFORMANCE NINTH EDITION
• Fetch: Read the next instruction from memory into the processor.
• Execute: Interpret the opcode and perform the indicated operation.
• Interrupt: If interrupts are enabled and an interrupt has occurred, save the
current process state and service the interrupt.
INSTRUCTION CYCLE的更多相关文章
- 高效的两段式循环缓冲区──BipBuffer
Simon Cooke,美国 (原作者) 北京理工大学 20981 陈罡(翻译) 写在前面的话: 循环缓冲区是一个非常常用的数据存储结构,已经被广泛地用于连续.流数据的存储和通信应用中.对于循环缓冲区 ...
- The Bip Buffer - The Circular Buffer with a Twist
Introduction The Bip-Buffer is like a circular buffer, but slightly different. Instead of keeping on ...
- sharc dsp 学习记录1---2014-07-30
从今天开始记录学习sharc dsp过程中的点点滴滴吧. DPI:Digital Peripheral Interface DAI:Digital Audio Interface SHARC ...
- OS面试题(转载)
转载自:http://placement.freshersworld.com/power-preparation/technical-interview-preparation/os-intervie ...
- SAE J1850 VPW Implement
---恢复内容开始--- OBDII Interface Project When I can ever find enough time away from schoolwork, I try to ...
- Software UART, Timer, PWM, External Interrupt
How can you add extra hardware UARTs to a 32bit TMS470 ARM7-based microcontroller at zero cost? Solu ...
- A brief introduction to Hashing and Rehashing
偶然发现一篇哈希的综述文章,虽然是1996年写的,里面的一些评测在今天看来早已经小case了.但是今天仍然极具参考价值. 地址:http://www.drdobbs.com/database/hash ...
- PatentTips - Write Combining Buffer for Sequentially Addressed Partial Line Operations
SUMMARY OF THE INVENTION The present invention pertains to a write combining buffer for use in a mic ...
- 时钟周期 VS 机器周期
时钟周期vs机器周期 Clock cycle The speed of a computer processor, or CPU, is determined by the clock cycle, ...
随机推荐
- HDU4871 Shortest-path tree(最短路径树 + 树的点分治)
题目大概要先求一张边有权的图的根为1的最短路径树,要满足根到各点路径序列的字典序最小:然后求这棵最短路径树包含k个结点的最长路径的长度和个数. 首先先构造出这棵字典序最小的最短路径树..好吧,我太傻逼 ...
- oracle去重
oracle去重 create table tmp_table3 as (SELECT seqno FROM (SELECT t.seqno,ROWID, ROW_NUMBER() OVER(PART ...
- Pick-up sticks[HDU1147]
Pick-up sticksTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...
- BZOJ3068 : 小白树
枚举每条树边,将其断开,那么两侧肯定取带权重心最优. 考虑如何求出每个子树的重心,枚举其所有儿子,通过重量关系就可以判断出重心位于哪棵子树. 然后将那棵子树的重心暴力往上爬即可,因为每个点作为重心肯定 ...
- Hive内部表外部表转化分析(装)
link:http://anyoneking.com/archives/127hive表分为内部表和外部表.外部表在删除的时候并不会删除到hdfs中的文件,比较安全,所以对于重要的需要进行分析的日志建 ...
- Codeforces Round #352 (Div. 2) B - Different is Good
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to b ...
- NOIP欢乐模拟赛 T3 解题报告
3.小澳的葫芦 (calabash.cpp/c/pas) [题目描述] 小澳最喜欢的歌曲就是<葫芦娃>. 一日表演唱歌,他尽了洪荒之力,唱响心中圣歌. 随之,小澳进入了葫芦世界. 葫芦世界 ...
- [BZOJ2799][Poi2012]Salaries
2799: [Poi2012]Salaries Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 91 Solved: 54[Submit][Statu ...
- BZOJ1565: [NOI2009]植物大战僵尸
Description Input Output 仅包含一个整数,表示可以获得的最大能源收入.注意,你也可以选择不进行任何攻击,这样能源收入为0. Sample Input 3 2 10 0 20 0 ...
- Linux安装卸载查看vsftpd
Linux & vsftpd 相关的命令: 查看---rpm -qa | grep vsftpd 卸载---rpm -e vsftpd 安装---rpm -ivh /media/(在此tab ...