ARM Assembly SOS中的64位划分 [英] 64 bit division in ARM Assembly SOS

问题描述

我正在计算加在一起的16个64位数字的平均值，我认为我正确地完成了所有加法运算，但是现在我需要弄清楚如何将64位数字除以16，这样我就卡住了！非常感谢您的任何帮助。到目前为止，这是我的代码。

  tableSize EQU 16 
 sum EQU 0x40000000 
平均EQU 0x40000008 
 MOV r8，＃14 
 
 ADR r0，表
 LDR r9，= sum 
 LDR r10，=平均值
 
 LDR r1，[r0 ]，＃1; hi＃1 
 LDR r2，[r0]，＃1; lo＃1 
 SUM 
 SUB r8，r8，＃1 
 LDR r3，[r0 ]，＃1; hi＃2 
 LDR r4，[r0]，＃1; lo＃2 
 ADDS r5，r2，r4; lo 1 + lo 2设置标志
 ADC r6， r1，r3; hi 1 + hi 2 +携带
 MOV r1，r6 
 MOV r2，r5 
 CMP r8，0 
 BNE SUM 
 
 STR r1，[r9]，＃8 
 STR r2，[r9] 
平均
;卡在此处
 
 
表DCQ 0x0200200AD00236DD 
 DCQ 0x00003401AAC4D097 
 DCQ 0x000001102ACFF200 
 DCQ 0x00010AA0AD3C66DF 
 DCQ 0x0000FC3D76400CCB 
 DCQ 0x000090045ACDD097 
 DCQ 0x00000FF000004551 
 DCQ 0x00000030000C $ 340B 
 DCQ 0x000001102ACFF200 
 DCQ 0x00010AA0AD3C66DF 
 DCQ 0x1000FC3D76400CCB 
 DCQ 0x000090045ACDD097 
 DCQ 0x00000FF000004551 
 DCQ 0x00000000003C66DF 
  
 
解决方案
鉴于 r0 和 r1 ，可以按照以下说明将其除以16：
  lsl r2，r0，＃28 
 asr r0，r0，＃4 
 orr r1 ，r2，r1和lsr＃4 
  
总而言之，我们要做的就是将两者都转移减半，然后将 r0 的低四位放入 r1 的高四位。
 
 
 
要获得无符号除法，应使用 lsr 而不是 asr 。 
 
 
 
在两种情况下，结果都会四舍五入为负无穷大。要将结果四舍五入到最接近的整数，可以在除法之前将8加到该整数上。另外，可以向正无穷大的方向加15。
I am calculating the average of sixteen 64 bit numbers added together and I think that I have done all the addition correctly, but now I need to figure out how to divide a 64 bit number by 16 and I am stuck! Any help would be great thank you so much. Here is my code so far.
tableSize       EQU     16
sum             EQU     0x40000000
average         EQU     0x40000008
                MOV r8, #14

                ADR r0, table
                LDR r9, =sum
                LDR r10,=average

                LDR r1, [r0], #1    ;hi #1
                LDR r2, [r0], #1    ;lo #1
SUM
                SUB r8, r8, #1
                LDR r3, [r0], #1    ;hi #2
                LDR r4, [r0], #1    ;lo #2
                ADDS    r5, r2, r4  ;lo 1 + lo 2 set flags
                ADC r6, r1, r3  ;hi 1 + hi 2 + carry
                MOV r1, r6
                MOV r2, r5
                CMP r8, 0   
                BNE SUM

                STR r1, [r9], #8
                STR r2, [r9]
 average
                ;stuck here 


table           DCQ     0x0200200AD00236DD
                DCQ     0x00003401AAC4D097
                DCQ     0x000001102ACFF200
                DCQ     0x00010AA0AD3C66DF
                DCQ     0x0000FC3D76400CCB
                DCQ     0x000090045ACDD097
                DCQ     0x00000FF000004551
                DCQ     0x00000000003C66DF
                DCQ     0x1000200AD00236DD
                DCQ     0x00003401AAC4D097
                DCQ     0x000001102ACFF200
                DCQ     0x00010AA0AD3C66DF
                DCQ     0x1000FC3D76400CCB
                DCQ     0x000090045ACDD097
                DCQ     0x00000FF000004551
                DCQ     0x00000000003C66DF

解决方案
Given that there's a 64 bit signed integer in r0 and r1, one can divide it by 16 with the following instructions:
    lsl     r2, r0, #28
    asr     r0, r0, #4
    orr     r1, r2, r1, lsr #4
In a nutshell, all we need to do is to shift both halves by four and put lower four bits of r0 into four upper bits of r1.

To get unsigned division, one should use lsr instead of asr. 

In both cases the result will be rounded towards minus infinity. To round the result towards the nearest integer one can add 8 to the integer before division. Also, one can add 15 to round towards plus infinity.

                        
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