在MIPS汇编中使用逻辑移位进行乘法 [英] Multiplication using Logical shifts in MIPS assembly

问题描述

有人可以给我一些指导，告诉我如何制作使用MIPS汇编中的移位倍增的代码吗?我不明白拥有2 ^ n的数字如何帮助我使用奇数被乘数进行乘法

Can someone please give me pointers on how I can go about making a code that multiplies using shifts in MIPS assembly? I don't understand how having a number 2^n can help me multiply using an odd multiplicand

我目前有此代码，我正在尝试制作一个计算器

I currently have this code, I'm trying to make a calculator

.text

li  $v0, 4 
la  $a0, ask_1
syscall

li  $v0,5
syscall
move    $s1, $v0


li  $v0, 4
la  $a0, ask_2
syscall

li  $v0,5
syscall
move    $s2, $v0

#sll    $s2, $s2, 3     #$s2 * $s2^3 = result
srl $s2, $s2, 1

li  $v0, 1
la  $a0, ($s2)
syscall


.data

ask_1:  .asciiz     "Enter Multiplier\n"
ask_2:  .asciiz     "Enter Multiplicand\n"
result: .asciiz         "The Answer is:\n"

推荐答案

将数字左移n位将数字乘以2 ⁿ .例如n << 3 = n*2^3 = n*8.相应的说明是

Shifting a number n bits left multiples the number by 2ⁿ. For example n << 3 = n*2^3 = n*8. The corresponding instruction is

SLL $s1, $s2, 1

要乘以任何数字，您可以将数字分成2s的幂.例如:

To multiply any number you can split the number into sum of power of 2s. For example:

n * 10 = n * 8 + n * 2 = n< 3 + n<< 1

n*10 = n*8 + n*2 = n << 3 + n << 1

SLL $t1, $s2, 1
SLL $t2, $s2, 3
ADD $s2, $t1, $t2

如果速度更快，您也可以使用减法

You can also use a subtraction if it's faster

n * 15 = n * 16-n = n<< 4-n

n*15 = n*16 - n = n << 4 - n

SLL $t1, $s2, 4
SUB $s1, $t1, $s2

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