Assembly 展开y86循环
我试图在y86代码中展开一个循环,但当我试图运行一个测试程序时,我得到了两个不同的值。注册。代码是:Assembly 展开y86循环,assembly,loop-unrolling,y86,Assembly,Loop Unrolling,Y86,我试图在y86代码中展开一个循环,但当我试图运行一个测试程序时,我得到了两个不同的值。注册。代码是: xorq %rax,%rax # count = 0; andq %rdx,%rdx # len <= 0? jle Done # if so, goto Done: Loop: mrmovq (%rdi), %r10 # read val from src... rmmovq %r10, (%rsi) #
xorq %rax,%rax # count = 0;
andq %rdx,%rdx # len <= 0?
jle Done # if so, goto Done:
Loop:
mrmovq (%rdi), %r10 # read val from src...
rmmovq %r10, (%rsi) # ...and store it to dst
andq %r10, %r10 # val <= 0?
jle Npos # if so, goto Npos:
#irmovq $1, %r10
#addq %r10, %rax
iaddq $1, %rax # count++
Npos:
irmovq $1, %r10
subq %r10, %rdx # len--
#irmovq $8, %r10
#addq %r10, %rdi
#addq %r10, %rsi
iaddq $8, %rdi # src++
iaddq $8, %rsi # dst++
andq %rdx,%rdx # len > 0?
jg Loop # if so, goto Loop:
Done:
ret
xorq%rax,%rax#count=0;
我刚把它修好了。我想在开始循环之前减小%rdx以正确展开函数
当您不知道迭代计数是展开因子的倍数时,您需要展开do{},而(--i>=0)将>编码为类似
--i;do{}while(i-=2>=0)代码>以确保您不会超调
xorq %rax,%rax # count = 0;
andq %rdx,%rdx # len <= 0?
jle Done # if so, goto Done:
Loop:
mrmovq (%rdi), %r10 # read val from src…
mrmovq 8(%rdi), %r11 # <- from class get second value
rmmovq %r10, (%rsi) # ...and store it to dst
rmmovq %r11, 8(%rsi) # store second val to dst
andq %r10, %r10 # val <= 0?
jle Npos # if so, goto Npos:
iaddq $1, %rax
Npos:
andq %r11, %r11 # check if src[1] <= 0
jle Npos2 # if it is, don’t increase count
iaddq $1, %rax
Npos2:
irmvoq %2, %r10
iaddq $16, %rdi # increase stack or base pointer to get next 2 vals
iaddq $16, %rsi # increase stack or base pointer to store next 2 vals
subq %r10, %rdx # decrease length by 2
jge Loop # go back into loop if length >= 2
len_cleanup:
iaddq $2, %rdx
cleanup:
irmovq $1, %r10
subq %r10, %rdx
jl Done # if length < 0, jmp to Done, no cleanup needed
mrmovq (%rdi), %r10 # get next val
rmmovq %r10, (%rsi) # move val onto stack
andq %r10, %r10 # check if val <= 0
jle Done # skip count if val < 0
iaddq $1, %rax # same as iaddq $1, %rax
Done:
ret