Assembler Code       Content of ROM
MOV @R10+,0(R11)       MOV @R10+,0(R11)

Length:   One or two words

Operation:   

  Move the contents of the source address (contents of R10) to the destination address (contents of R11). Register R10 is incremented by 1 for a byte operation, or 2 for a word operation after the fetch; it points to the next address without any overhead. This is useful for table processing.

Comment:

   Valid only for source operand. The substitute for destination operand is 0(Rd) plus second instruction INCD Rd.

Example:   MOV @R10+,0(R11)

Assembly之instruction之Indirect Autoincrement Mode的更多相关文章

  1. Assembly之instruction之JC

    JC Jump if carry setJHS  Jump if higher or same Syntax JC label JHS label Operation If C = 1: PC + 2 ...

  2. Assembly之instruction之JUMP

    JMP  Jump unconditionally Syntax   JMP  label Operation PC + 2 × offset −> PC Description The 10- ...

  3. Assembly之Instruction之Byte and Word

    Byte and word issues The MSP430 is byte-addressed, and little-endian. Word operands must be located ...

  4. Assembly之instruction之Status register

    The status register (SR/R2), used as a source or destination register, can be used in the register m ...

  5. Assembly之instruction之Register Mode

    Assembler Code Content of ROM MOV R10,R11 MOV R10,R11 Length: One or two words Operation: Move the c ...

  6. Assembly之instruction之MOV

    MOV[.W]   Move source to destinationMOV.B Move source to destination Syntax MOV  src,dst  or       M ...

  7. Assembly之instruction之CMP

    CMP[.W]  Compare source and destinationCMP.B  Compare source and destination Syntax  CMP src,dst or ...

  8. BitHacks

    备份文件时看到的.我以前居然下过这东西. 2016-12-4 12:05:52更新 纯文本格式真棒.假如使用word写的我能拷过来格式还不乱?? Markdown真好. Bit Hacks By Se ...

  9. Bit Twiddling Hacks

    http://graphics.stanford.edu/~seander/bithacks.html Bit Twiddling Hacks By Sean Eron Andersonseander ...

随机推荐

  1. 第一节:web爬虫之requests

    Requests库是用Python编写的,并且Requests是一个优雅而简单的Python HTTP库,在使用Requests库时更加方便,可以节约我们大量的工作,完全满足HTTP测试需求.

  2. 解析 XML 数据

    在几个月前我有做过这样的记录,其目的是避免解析 XML 时手工编写太多的代码,造成重复的体力劳动.后来经过一番资料的查找,我发现其实并没有必要做这样的工具,因为 C# 已经为我们提供了更好的解决方案了 ...

  3. 【Codeforces Global Round 1 A】Parity

    [链接] 我是链接,点我呀:) [题意] 给你一个k位数b进制的进制转换. 让你求出来转成10进制之后这个数字是奇数还是偶数 [题解] 模拟一下转换的过程,加乘的时候都记得对2取余就好 [代码] im ...

  4. 【ACM】nyoj_6_喷水装置(1)_201308150853

    喷水装置(一)时间限制:3000 ms  |  内存限制:65535 KB 难度:3描述 现有一块草坪,长为20米,宽为2米,要在横中心线上放置半径为Ri的喷水装置,每个喷水装置的效果都会让以它为中心 ...

  5. POJ 3304 segments 线段和直线相交

    Segments Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14178   Accepted: 4521 Descrip ...

  6. J - Simpsons’ Hidden Talents

    Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had. Marg ...

  7. [bzoj1634][Usaco2007 Jan]Protecting the Flowers 护花_贪心

    Protecting the Flowers 护花 bzoj-1634 Usaco-2007 Jan 题目大意:n头牛,每头牛有两个参数t和atk.表示弄走这头牛需要2*t秒,这头牛每秒会啃食atk朵 ...

  8. c++实现数据结构1.顺序表

    头文件seqlist.h #ifndef _SEQLIST_H_ #define _SEQLIST_H_ #include<iostream> using namespace std; t ...

  9. [IOS/翻译]Core Services Layer

    本文是本人自己辛苦翻译的,请转载的朋友注明,翻译于Z.MJun的CSDN的博客 http://blog.csdn.net/Zheng_Paul,感谢. 翻译于2015年10月4日 Core Servi ...

  10. ubuntu网卡ip的配置

    ifconfig 命令的结果 和 ip addr (或者查看具体网卡的是 ip addr show eth0) 看到的结果不一样, ip addr show eth0 可以看到eth0网卡上面的多个 ...