A typical implementation
Booth's algorithm can be implemented by repeatedly adding (with ordinary unsigned binary addition) one of two predetermined values A and S to a product P, then performing a rightward arithmetic shift on P. Let m and r be the multiplicand and multiplier, respectively; and let x and y represent the number of bits in m and r.
Determine the values of A and S, and the initial value of P. All of these numbers should have a length equal to (x + y + 1).
A: Fill the most significant (leftmost) bits with the value of m. Fill the remaining (y + 1) bits with zeros.
S: Fill the most significant bits with the value of (−m) in two's complement notation. Fill the remaining (y + 1) bits with zeros.
P: Fill the most significant x bits with zeros. To the right of this, append the value of r. Fill the least significant (rightmost) bit with a zero.
Determine the two least significant (rightmost) bits of P.
If they are 01, find the value of P + A. Ignore any overflow.
If they are 10, find the value of P + S. Ignore any overflow.
If they are 00, do nothing. Use P directly in the next step.
If they are 11, do nothing. Use P directly in the next step.
Arithmetically shift the value obtained in the 2nd step by a single place to the right. Let P now equal this new value.
Repeat steps 2 and 3 until they have been done y times.
Drop the least significant (rightmost) bit from P. This is the product of m and r.

For more details about Booth algorithm,refer to here

######################################################################################################
# Program: Booth Multiplication Algorithm
# Language: MIPS Assembly (32-bit)
# Arguments:
#   $2 stores multiplicand (m)
#   $3 stores multiplier   (r)
#   $5 stores the number of bits of each element
#   $6 stores high 32-bit of result
#   $7 stores low  32-bit of result
# Author: brant-ruan
# Date: 2016-03-11
# IDE: MARS 4.5
######################################################################################################
.text
    add $2, $0, -8      # multiplicand (m)
    add $3, $0, 3       # multiplier   (r)
    xor $4, $4, $4      # i = 0
    add $5, $0, 32      # max_iteration times
    xor $6, $6, $6      # high 32-bit of P
    add $7, $0, $3      # low  32-bit of P
    xor $9, $9, $9      # store the rightmost bit of P
begin:
    sll $10, $7, 1      # $10 = $7 << 1
    or  $9, $9, $10     #
    and $9, $9, 3       # generate the 2 rightmost bits of P
    beq $9, 0, shift    # 0 then goto shift
    beq $9, 3, shift    # 3 then goto shift
    beq $9, 2, case_2   # 2 then goto case_2
case_1:             # P(high 32-bit) = P(high 32-bit) + m
    add $6, $6, $2
    j shift
case_2:             # P(high 32-bit) = P(high 32-bit) - m
    sub $6, $6, $2
shift:
    and $8, $6, 1       # $8 stores the rightmost bit of high 32-bit of P
    sra $6, $6, 1       # high 32-bit >> 1
    sll $8, $8, 31      #
    and $9, $7, 1       # $9 stores the rightmost bit of P
    srl $7, $7, 1       # logical shift ! Not sra !
    or  $7, $8, $7      # rightmost of high 32-bit becomes the leftmost of low 32-bit
    add $4, $4, 1       # i++
    bne $4, $5, begin   # if i != 32, back to begin
end:

Booth Multiplication Algorithm [ASM-MIPS]的更多相关文章

  1. Design and Analysis of Algorithms_Fundamentals of the Analysis of Algorithm Efficiency

    I collect and make up this pseudocode from the book: <<Introduction to the Design and Analysis ...

  2. [转]Whirlwind Tour of ARM Assembly

    ref:http://www.coranac.com/tonc/text/asm.htm 23.1. Introduction Very broadly speaking, you can divid ...

  3. MARS3.6 Programming

    An Assembly Language I.D.E. To Engage Students Of All Levels * A Tutorial *2007 CCSC: Central Plains ...

  4. Conquer and Divide经典例子之Strassen算法解决大型矩阵的相乘

    在通过汉诺塔问题理解递归的精髓中我讲解了怎么把一个复杂的问题一步步recursively划分了成简单显而易见的小问题.其实这个解决问题的思路就是算法中常用的divide and conquer, 这篇 ...

  5. Codeforces 980E The Number Games - 贪心 - 树状数组

    题目传送门 传送点I 传送点II 传送点III 题目大意 给定一颗有$n$个点的树,$i$号点的权值是$2^{i}$要求删去$k$个点,使得剩下的点仍然连通,并且总权值和最大,问删去的所有点的编号. ...

  6. Strassen优化矩阵乘法(复杂度O(n^lg7))

    按照算法导论写的 还没有测试复杂度到底怎么样 不过这个真的很卡内存,挖个坑,以后写空间优化 还有Matthew Anderson, Siddharth Barman写了一个关于矩阵乘法的论文 < ...

  7. 各种字符串Hash函数(转)

    /// @brief BKDR Hash Function /// @detail 本 算法由于在Brian Kernighan与Dennis Ritchie的<The C Programmin ...

  8. memory ordering 内存排序

    Memory ordering - Wikipedia https://en.wikipedia.org/wiki/Memory_ordering https://zh.wikipedia.org/w ...

  9. 布斯乘法 Mips实现 - Booth Algorithm

    看了很久网上没有现成的代码和好一点的图,因此当一回搬运工.转自stackoverflow 布斯乘法器的Mips实现方法: .data promptStart: .asciiz "This p ...

随机推荐

  1. 如何在window Form中使用Font Awesome?

    随着技术的发展,web上以前的图片按钮现在逐步换成了图标字体,这些图标字体是矢量的,矢量图意味着每个图标都能在所有大小的屏幕上完美呈现,可以随时更改大小和颜色,而且不失真,真心给人一种“高大上”的感觉 ...

  2. CSV.js – 用于 CSV 解析和编码的 JS 工具库

    逗号分隔值(CSV )文件用于以以纯文本的形式存储表格化数据(数字和文本). CSV 文件包含任意数量的记录,通过某种换行符分隔,每条记录由字段,其他一些字符或字符串分隔,最常用的是文字逗号或制表符. ...

  3. go语言选择语句 switch case

    根据传入条件的不同,选择语句会执行不同的语句.下面的例子根据传入的整型变量i的不同而打印不同的内容: switch i { case 0: fmt.Printf("0") case ...

  4. scroll事件实现监控滚动条并分页显示示例(zepto.js)

    scroll事件实现监控滚动条并分页显示示例(zepto.js  ) 需求:在APP落地页上的底部位置显示此前其他用户的购买记录,要求此div盒子只显示3条半,但一页有10条,div内的滑动条滑到一页 ...

  5. HTML5拖放(drag and drop)与plupload的懒人上传

    HTML5拖放能够将本地的文件拖放到页面上,plupload又是很好的文件上传插件,而今天就将两者结合,做了个文件拖拽上传的功能. 简述HTML5拖放 拖放是HTML5标准的一部分,任何元素都能够拖放 ...

  6. Asp.net web hosting

      start /D "C:\Program Files\Common Files\Microsoft Shared\DevServer\10.0" /B WebDev.WebSe ...

  7. JavaScript学习02 基础语法

    JavaScript学习02 基础语法 JavaScript中很多基础内容和Java中大体上基本一样,所以不需要再单独重复讲了,包括: 各种算术运算符.比较运算符.逻辑运算符: if else语句.s ...

  8. iOS 获取emoji表情和拦截emoji表情

      1 2 //将数字转为 #define EMOJI_CODE_TO_SYMBOL(x) ((((0x808080F0 | (x & 0x3F000) >> 4) | (x &a ...

  9. iOS拨打电话的三种方式

    iOS拨打电话的三种方式 1,这种方法,拨打完电话回不到原来的应用,会停留在通讯录里,而且是直接拨打,不弹出提示 1 2 var string = "tel:" + "1 ...

  10. 我曾经的第一个OC程序

    一. OC简介 C语言的基础上,增加了一层最小的面向对象语法 完全兼容C语言 可以在OC代码中混入C语言代码,甚至是C++代码 可以使用OC开发Mac OS X平台和iOS平台的应用程序 二. OC语 ...