MIPS中三个整数的乘法 [英] Multiplication of three integers in MIPS

问题描述

我想在MIPS中乘以三个整数.我的第一个想法是将第一个和第二个相乘，然后将结果与第三个相乘(就像我对add所做的那样).但是结果在64位的HI和LOW中给出.那么如何将其乘以第三个因子呢?

I want to multiply three integers in MIPS. My first thought was to multiply the first and the second one, and after that the result with the third one (like I did it with add). But the result is given in HI and LOW in 64-bit. So how can I multiply it with the third factor?

和: 32位整数* 32位整数= 64位整数.理论上会得到什么:

And: 32-bit integer * 32-bit integer = 64-bit integer. What (in the theory) would give that:

32位整数* 32位整数* 32位整数= ??

96位? 128?

感谢您的提示.

推荐答案

将n位数字与m位数字相乘会生成(n + m)位数字. 因此，将三个32位数字相乘可得出96位乘积

Multiplying an n-bit number with an m-bit number produces a (n+m) bit number. So multiplying three 32-bit numbers produces a 96-bit product

假设三个数字分别是a，b和c，我们得到以下结果

Suppose the three numbers are a, b and c, we have the following result

  a*b  =   ABH*2³² + ABL
c(a*b) = c(ABH*2³² + ABL)
       = c*ABH*2³² + c*ABL
       = (CABH_H*2³² + CABH_L)*2³² + (CABL_H*2³² + CABL_L)
       = CABH_H*2⁶⁴ + (CABH_L + CABL_H)*2³² + CABL_L

其中，H和L是结果的高部分和低部分，分别存储在HI和LO寄存器中.在这里，将2 ³² 和2 ⁶⁴ 的乘法分别替换为左移32和64.

where H and L are the high and low parts of the result which are stored in the HI and LO registers respectively. Here the multiplications by 2³² and 2⁶⁴ will be replaced by shifting left by 32 and 64 respectively.

因此在MIPS32中，如果a，b和c存储在$ s0，$ s1和$ s2中，我们可以进行如下计算

So in MIPS32 if a, b and c are stored in $s0, $s1 and $s2 we can do the math as below

multu   $s0, $s1        # (HI, LO) = a*b
mfhi    $t0             # t0 = ABH
mflo    $t1             # t1 = ABL

multu   $s2, $t1        # (HI, LO) = c*ABL
mfhi    $t4             # t4 = CABL_H
mflo    $s7             # s7 = CABL_L

multu   $s2, $t0        # (HI, LO) = c*ABH
mfhi    $s5             # s5 = CABH_H
mflo    $t3             # t3 = CABH_L

addu    $s6, $t3, $t4   # s6 = CABH_L + CABL_H
sltu    $t5, $s6, $t3   # carry if s6 overflows
addu    $s5, $s5, $t5   # add carry to s5
# result = (s5 << 64) | (s6 << 32) | s7

结果存储在元组($ s5，$ s6，$ s7)中

The result is stored in the tuple ($s5, $s6, $s7)

在MIPS64中，事情要简单得多

Things will be much simpler in MIPS64:

mul(unsigned int, unsigned int, unsigned int):
        multu   $4,$5
        dext    $6,$6,0,32
        mfhi    $3
        mflo    $2
        dins    $2,$3,32,32
        dmultu  $6,$2
        mflo    $3
        j       $31
        mfhi    $2

这是您可能需要对签名的操作进行一些修改

You may need some slight modification for signed operations

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