基本上，我需要通过减小整体代码的大小来减小内存大小，或者提高其运行效率，从而使这段代码更有效率。我正在使用Thumb 2以及Cortex-M3。
我尝试减少使用的 MOV 函数的数量，但虽然这确实减少了整体代码大小，但由于代码的工作方式，它需要每个单独的部分来获取并将结果存储在寄存器中，所以我有点困惑如何改进它。代码目前处于默认状态。
THUMB
AREA RESET, CODE, READONLY
EXPORT __Vectors
EXPORT Reset_Handler
__Vectors
DCD 0x00180000 ; top of the stack
DCD Reset_Handler ; reset vector - where the program starts
AREA 2a_Code, CODE, READONLY
Reset_Handler
ENTRY
num_words EQU (end_source-source)/4 ; number of words to copy
start
LDR r0,=source ; point to the start of the area of memory to copy from
LDR r1,=dest ; point to the start of the area of memory to copy to
MOV r2,#num_words ; get the number of words to copy
; find out how many blocks of 8 words need to be copied - it is assumed
; that it is faster to load 8 data items at a time, rather than load
; individually
block
MOVS r3,r2,LSR #3 ; find the number of blocks of 8 words
BEQ individ ; if no blocks to copy, just copy individual words
; copy and process blocks of 8 words
block_loop
LDMIA r0!,{r5-r12} ; get 8 words to copy as a block
MOV r4,r5 ; get first item
BL data_processing ; process first item
MOV r5,r4 ; keep first item
MOV r4,r6 ; get second item
BL data_processing ; process second item
MOV r6,r4 ; keep second item
MOV r4,r7 ; get third item
BL data_processing ; process third item
MOV r7,r4 ; keep third item
MOV r4,r8 ; get fourth item
BL data_processing ; process fourth item
MOV r8,r4 ; keep fourth item
MOV r4,r9 ; get fifth item
BL data_processing ; process fifth item
MOV r9,r4 ; keep fifth item
MOV r4,r10 ; get sixth item
BL data_processing ; process sixth item
MOV r10,r4 ; keep sixth item
MOV r4,r11 ; get seventh item
BL data_processing ; process seventh item
MOV r11,r4 ; keep seventh item
MOV r4,r12 ; get eighth item
BL data_processing ; process eighth item
MOV r12,r4 ; keep eighth item
STMIA r1!,{r5-r12} ; copy the 8 words
SUBS r3,r3,#1 ; move on to the next block
BNE block_loop ; continue until last block reached
; there may now be some data items available (fewer than 8)
; find out how many of these individual words need to be copied
individ
ANDS r3,r2,#7 ; find the number of words that remain to copy individually
BEQ exit ; skip individual copying if none remains
; copy the excess of words
individ_loop
LDR r4,[r0],#4 ; get next word to copy
BL data_processing ; process the item read
STR r4,[r1],#4 ; copy the word
SUBS r3,r3,#1 ; move on to the next word
BNE individ_loop ; continue until the last word reached
; languish in an endless loop once all is done
exit
B exit
; subroutine to scale a value by 0.5 and then saturate values to a maximum of 5
data_processing
CMP r4,#10 ; check whether saturation is needed
BLT divide_by_two ; if not, just divide by 2
MOV r4,#5 ; saturate to 5
BX lr
divide_by_two
MOV r4,r4,LSR #1 ; perform scaling
BX lr
AREA 2a_ROData, DATA, READONLY
source ; some data to copy
DCD 1,2,3,4,5,6,7,8,9,10,11,0,4,6,12,15,13,8,5,4,3,2,1,6,23,11,9,10
end_source
AREA 2a_RWData, DATA, READWRITE
dest ; copy to this area of memory
SPACE end_source-source
end_dest
END
基本上，我需要代码将结果存储在每个寄存器中，同时减小大小或使其执行速度更快。感谢您的帮助。
这是主循环的稍微优化版本。 鉴于您正在为 Cortex M3 编程，因此不可能进行超标量或 SIMD 处理，因为您的 CPU 不支持它。 这和你的代码之间的主要区别是：
- 所有相关功能都内联
- 逻辑已优化了一点
- 无用的移动指令已被省略
此代码在每个表条目中运行 10 个周期，外加一些初始分支的指令以及最终的分支错误预测。
.syntax unified
.thumb
@ r0: source
@ r1: destination
@ r2: number of words to copy
@ the number in front of the comment is the number
@ of cycles needed to execute the instruction
block: cbz r2, .Lbxlr @ 2 return if nothing to copy
.Loop: ldmia r0!, {r3} @ 2 load one item from source
cmp r3, #10 @ 1 need to scale?
ite lt @ 1 if r3 < 10:
lsrlt r3, r3, #1 @ 1 then r3 >>= 1
movge r3, #5 @ 1 else r3 = 5
stmia r1!, {r3} @ 2 store to destination
subs r2, r2, #1 @ 1 decrement #words
bne .Loop @ 1 continue if not done yet
.Lbxlr: bx lr
您可以通过展开循环一次，将两个条目的周期减少到 16 个周期(每个条目 8 个周期)。 请注意，这几乎使代码长度增加了三倍，而性能却很小。
.syntax unified
.thumb
@ r0: source
@ r1: destination
@ r2: number of words to copy
@ the number in front of the comment is the number
@ of cycles needed to execute the instruction
@ first check if the number of elements is even or odd
@ leave this out if it's know to be even
block: tst r2, #1 @ 1 odd number of entries to copy?
beq .Leven @ 2 if not, proceed with eveness check
ldmia r0!, {r3} @ 2 load one item from source
cmp r3, #10 @ 1 need to scale?
ite lt @ 1 if r3 < 10:
lsrlt r3, r3, #1 @ 1 then r3 >>= 1
movge r3, #5 @ 1 else r3 = 5
stmia r1!, {r3} @ 2 store to destination
subs r2, r2, #1 @ 1 decrement #words
@ check if any elements are left
@ remove if you know that at least two elements are present
.Leven: cbz r2, .Lbxlr @ 2 return if no entries left.
.Loop: ldmia r0!, {r3, r4} @ 3 load two items from source
cmp r3, #10 @ 1 need to scale?
ite lt @ 1 if r3 < 10:
lsrlt r3, r3, #1 @ 1 then r3 >>= 1
movge r3, #5 @ 1 else r3 = 5
cmp r4, #10 @ 1 need to scale?
ite lt @ 1 if r5 < 10:
lsrlt r4, r4, #1 @ 1 then r4 >>= 1
movge r4, #5 @ 1 else r4 = 5
stmia r1!, {r3, r4} @ 3 store to destination
subs r2, r2, #2 @ 1 decrement #words twice
bne .Loop @ 1 continue if not done yet
.Lbxlr: bx lr
每个元素可以通过四次展开循环来实现 7 个周期，但我认为这太多了。
请注意，这段代码在 GNU 中作为语法。 为您的汇编程序修改它应该是微不足道的。