问题
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
回答1:
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.
来源：https://stackoverflow.com/questions/26633800/64-bit-division-in-arm-assembly-sos