请教ARM中的预取命令PLD的使用
我现在在看android2.3.3提供的关于ARM平台的memcmp这个函数的实现代码,它是用汇编编写的,如下:
.text
.global __memcmp16
.type __memcmp16, %function
.align 4
/*
* Optimized memcmp16() for ARM9.
* This would not be optimal on XScale or ARM11, where more prefetching
* and use of PLD will be needed.
* The 2 major optimzations here are
* (1) The main loop compares 16 bytes at a time
* (2) The loads are scheduled in a way they won't stall
*/
__memcmp16:
.fnstart
PLD (r0, #0)
PLD (r1, #0)
/* take of the case where length is nul or the buffers are the same */
cmp r0, r1
cmpne r2, #0
moveq r0, #0
bxeq lr
/* since r0 hold the result, move the first source
* pointer somewhere else
*/
mov r3, r0
/* make sure we have at least 12 words, this simplify things below
* and avoid some overhead for small blocks
*/
cmp r2, #12
bpl 0f
/* small blocks (less then 12 words) */
PLD (r0, #32)
PLD (r1, #32)
1: ldrh r0, [r3], #2
ldrh ip, [r1], #2
subs r0, r0, ip
bxne lr
subs r2, r2, #1
bne 1b
bx lr
.save {r4, lr}
/* save registers */
0: stmfd sp!, {r4, lr}
/* align first pointer to word boundary */
tst r3, #2
beq 0f
ldrh r0, [r3], #2
ldrh ip, [r1], #2
sub r2, r2, #1
subs r0, r0, ip
/* restore registers and return */
ldmnefd sp!, {r4, lr}
bxne lr
.fnend
0: /* here the first pointer is aligned, and we have at least 3 words
* to process.
*/
/* see if the pointers are congruent */
eor r0, r3, r1
ands r0, r0, #2
bne 5f
/* congruent case, 16 half-words per iteration
* We need to make sure there are at least 16+2 words left
* because we effectively read ahead one long word, and we could
* read past the buffer (and segfault) if we're not careful.
*/
ldr ip, [r1]
subs r2, r2, #(16 + 2)
bmi 1f
0: ///PLD是宏定义,意思是ARM指令集支持pld命令,就用pld,否则为空
///我不明白的是:这里为什么要把r3+64的地址中的数据取出来,这里好像也没有用到r3+64这个地址中的数据啊????? 请各位忙吧分析下,谢谢了,
PLD (r3, #64)
PLD (r1, #64) ///这里为什么要加64?????
ldr r0, [r3], #4
ldr lr, [r1, #4]!
eors r0, r0, ip
ldreq r0, [r3], #4
ldreq ip, [r1, #4]!
eoreqs r0, r0, lr
ldreq r0, [r3], #4
ldreq lr, [r1, #4]!
eoreqs r0, r0, ip
ldreq r0, [r3], #4
ldreq ip, [r1, #4]!
eoreqs r0, r0, lr
ldreq r0, [r3], #4
ldreq lr, [r1, #4]!
eoreqs r0, r0, ip
ldreq r0, [r3], #4
ldreq ip, [r1, #4]!
eoreqs r0, r0, lr
ldreq r0, [r3], #4
ldreq lr, [r1, #4]!
eoreqs r0, r0, ip
ldreq r0, [r3], #4
ldreq ip, [r1, #4]!
eoreqs r0, r0, lr
bne 2f
subs r2, r2, #16
bhs 0b
/* do we have at least 2 words left? */
1: adds r2, r2, #(16 - 2 + 2)
bmi 4f
/* finish off 2 words at a time */
3: ldr r0, [r3], #4
ldr ip, [r1], #4
eors r0, r0, ip
bne 2f
subs r2, r2, #2
bhs 3b
/* are we done? */
4: adds r2, r2, #2
bne 8f
/* restore registers and return */
mov r0, #0
ldmfd sp!, {r4, lr}
bx lr
2: /* the last 2 words are different, restart them */
ldrh r0, [r3, #-4]
ldrh ip, [r1, #-4]
subs r0, r0, ip
ldreqh r0, [r3, #-2]
ldreqh ip, [r1, #-2]
subeqs r0, r0, ip
/* restore registers and return */
ldmfd sp!, {r4, lr}
bx lr
/* process the last few words */
8: ldrh r0, [r3], #2
ldrh ip, [r1], #2
subs r0, r0, ip
bne 9f
subs r2, r2, #1
bne 8b
9: /* restore registers and return */
ldmfd sp!, {r4, lr}
bx lr
5: /*************** non-congruent case ***************/
/* align the unaligned pointer */
bic r1, r1, #3
ldr lr, [r1], #4
sub r2, r2, #8
6:
PLD (r3, #64)
PLD (r1, #64)
mov ip, lr, lsr #16
ldr lr, [r1], #4
ldr r0, [r3], #4
orr ip, ip, lr, lsl #16
eors r0, r0, ip
moveq ip, lr, lsr #16
ldreq lr, [r1], #4
ldreq r0, [r3], #4
orreq ip, ip, lr, lsl #16
eoreqs r0, r0, ip
moveq ip, lr, lsr #16
ldreq lr, [r1], #4
ldreq r0, [r3], #4
orreq ip, ip, lr, lsl #16
eoreqs r0, r0, ip
moveq ip, lr, lsr #16
ldreq lr, [r1], #4
ldreq r0, [r3], #4
orreq ip, ip, lr, lsl #16
eoreqs r0, r0, ip
bne 7f
subs r2, r2, #8
bhs 6b
sub r1, r1, #2
/* are we done? */
adds r2, r2, #8
moveq r0, #0
beq 9b
/* finish off the remaining bytes */
b 8b
7: /* fix up the 2 pointers and fallthrough... */
sub r1, r1, #2
b 2b
