* .macro restore_irqs, olddaif
* msrdaif, \olddaif
. * endm
*/
restore_irqs x9
//x1存储ccsidr_el1内容，低三位是(Log2(Number of bytes in cache line)) – 4
//加4后x2=(Log2(Numberof bytes in cache line))
and x2, x1, #7 // extract the length of the cachelines
add x2, x2, #4 // add 4 (line length offset)
mov x4, #0x3ff
//逻辑右移3位，提取bits[12:3](Associativityof cache) – 1，
//x4存储cache的way数
and x4, x4, x1, lsr #3 // find maximum number on the way size
//计算x4前面0的个数，存到x5
clz x5, x4 // find bit position of way sizeincrement
//提取bits[27:13]位：(Number of sets in cache) - 1
mov x7, #0x7fff
//x7中存储cache中的set数
and x7, x7, x1, lsr #13 // extract max number of the index size
loop2:
//把x4值备份
mov x9, x4 // create working copy of max waysize
loop3:
//把需要操作哪个way存储到x6
lsl x6, x9, x5
//确定操作哪一级的哪个way（x10指定操作哪一级cache）
orr x11, x10, x6 // factor way and cache number intox11
//确定操作哪个set
lsl x6, x7, x2
orr x11, x11, x6 // factor index number into x11
//x11中存储了哪一级cache（10），哪一路cache（x9），哪个set（x7）
dc cisw, x11 // clean & invalidate by set/way
//way数-1
subs x9, x9, #1 // decrementthe way
b.ge loop3
subs x7, x7, #1 // decrementthe index
b.ge loop2
skip:
add x10, x10, #2 // increment cache number，
//为什么加2不是1？见loop1标号处解释
cmp x3, x10
b.gt loop1
finished:
mov x10, #0 // swith back to cache level 0
msr csselr_el1, x10 // select current cache level incsselr
dsb sy
isb
ret
ENDPROC(__flush_dcache_all)
如果你对此有疑问，欢迎留言讨论。