当前位置: 技术问答>linux和unix
linux下如何在内核空间实现memcpy的汇编代码？
来源: 互联网 发布时间：2016-09-27
本文导语: 在内核空间的汇编memcpy代码如何实现呢？ 我是这样写的 9 memcpy: 10 push ebp 11 mov ebp, esp 12 push esi 13 push edi 14 push ...
在内核空间的汇编memcpy代码如何实现呢？
我是这样写的
我调用了之后就出现了page fault错误
也不知道哪儿出错了 还请高手指点一下
我是这样写的
9 memcpy:
10 push ebp
11 mov ebp, esp
12 push esi
13 push edi
14 push ecx
15
16 mov edi, [ebp + 8]
17 mov esi, [ebp + 12]
18 mov ecx, [ebp + 16]
19 .1:
20 cmp ecx, 0
21 jz .2
22 mov al, [ds:esi]
23 inc esi
24
25 mov byte [es:edi], al
26 inc edi
27 dec ecx
28 jmp .1
29 .2:
30 mov eax, [ebp + 8]
31 pop ecx
32 pop edi
33 pop esi
34 mov esp, ebp
35 pop ebp
36 ret
我调用了之后就出现了page fault错误
也不知道哪儿出错了 还请高手指点一下
|
int memcmp(const void *cs, const void *ct, size_t n)
{
register unsigned long r2 asm("2") = (unsigned long) cs;
register unsigned long r3 asm("3") = (unsigned long) n;
register unsigned long r4 asm("4") = (unsigned long) ct;
register unsigned long r5 asm("5") = (unsigned long) n;
int ret;
asm volatile ("0: clcle %1,%3,0n"
" jo 0bn"
" ipm %0n"
" srl %0,28"
: "=&d" (ret), "+a" (r2), "+a" (r3), "+a" (r4), "+a" (r5)
: : "cc" );
if (ret)
ret = *(char *) r2 - *(char *) r4;
return ret;
}
不知道找错没有，这也是在内核找到的。
|
void *memcpy(void *pdst,
const void *psrc,
size_t pn)
{
/* Ok. Now we want the parameters put in special registers.
Make sure the compiler is able to make something useful of this.
As it is now: r10 -> r13; r11 -> r11 (nop); r12 -> r12 (nop).
If gcc was allright, it really would need no temporaries, and no
stack space to save stuff on. */
register void *return_dst __asm__ ("r10") = pdst;
register char *dst __asm__ ("r13") = pdst;
register const char *src __asm__ ("r11") = psrc;
register int n __asm__ ("r12") = pn;
/* When src is aligned but not dst, this makes a few extra needless
cycles. I believe it would take as many to check that the
re-alignment was unnecessary. */
if (((unsigned long) dst & 3) != 0
/* Don't align if we wouldn't copy more than a few bytes; so we
don't have to check further for overflows. */
&& n >= 3)
{
if ((unsigned long) dst & 1)
{
n--;
*(char*)dst = *(char*)src;
src++;
dst++;
}
if ((unsigned long) dst & 2)
{
n -= 2;
*(short*)dst = *(short*)src;
src += 2;
dst += 2;
}
}
/* Decide which copying method to use. */
if (n >= 44*2) /* Break even between movem and
move16 is at 38.7*2, but modulo 44. */
{
/* For large copies we use 'movem' */
/* It is not optimal to tell the compiler about clobbering any
registers; that will move the saving/restoring of those registers
to the function prologue/epilogue, and make non-movem sizes
suboptimal.
This method is not foolproof; it assumes that the "asm reg"
declarations at the beginning of the function really are used
here (beware: they may be moved to temporary registers).
This way, we do not have to save/move the registers around into
temporaries; we can safely use them straight away.
If you want to check that the allocation was right; then
check the equalities in the first comment. It should say
"r13=r13, r11=r11, r12=r12" */
__asm__ volatile ("
;; Check that the following is true (same register names on
;; both sides of equal sign, as in r8=r8):
;; %0=r13, %1=r11, %2=r12
;;
;; Save the registers we'll use in the movem process
;; on the stack.
subq 11*4,$sp
movem $r10,[$sp]
;; Now we've got this:
;; r11 - src
;; r13 - dst
;; r12 - n
;; Update n for the first loop
subq 44,$r12
0:
movem [$r11+],$r10
subq 44,$r12
bge 0b
movem $r10,[$r13+]
addq 44,$r12 ;; compensate for last loop underflowing n
;; Restore registers from stack
movem [$sp+],$r10"
/* Outputs */ : "=r" (dst), "=r" (src), "=r" (n)
/* Inputs */ : "0" (dst), "1" (src), "2" (n));
}
/* Either we directly starts copying, using dword copying
in a loop, or we copy as much as possible with 'movem'
and then the last block (= 16 )
{
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
n -= 16;
}
/* A switch() is definitely the fastest although it takes a LOT of code.
* Particularly if you inline code this.
*/
switch (n)
{
case 0:
break;
case 1:
*(char*)dst = *(char*)src;
break;
case 2:
*(short*)dst = *(short*)src;
break;
case 3:
*((short*)dst)++ = *((short*)src)++;
*(char*)dst = *(char*)src;
break;
case 4:
*((long*)dst)++ = *((long*)src)++;
break;
case 5:
*((long*)dst)++ = *((long*)src)++;
*(char*)dst = *(char*)src;
break;
case 6:
*((long*)dst)++ = *((long*)src)++;
*(short*)dst = *(short*)src;
break;
case 7:
*((long*)dst)++ = *((long*)src)++;
*((short*)dst)++ = *((short*)src)++;
*(char*)dst = *(char*)src;
break;
case 8:
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
break;
case 9:
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*(char*)dst = *(char*)src;
break;
case 10:
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*(short*)dst = *(short*)src;
break;
case 11:
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*((short*)dst)++ = *((short*)src)++;
*(char*)dst = *(char*)src;
break;
case 12:
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
break;
case 13:
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*(char*)dst = *(char*)src;
break;
case 14:
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*(short*)dst = *(short*)src;
break;
case 15:
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*((long*)dst)++ = *((long*)src)++;
*((short*)dst)++ = *((short*)src)++;
*(char*)dst = *(char*)src;
break;
}
return return_dst; /* destination pointer. */
} /* memcpy() */
|
解决不了，不过帮你顶下
|
不会，关注，帮顶
您可能感兴趣的文章:
本站(WWW.)旨在分享和传播互联网科技相关的资讯和技术，将尽最大努力为读者提供更好的信息聚合和浏览方式。
本站(WWW.)站内文章除注明原创外，均为转载、整理或搜集自网络。欢迎任何形式的转载，转载请注明出处。
本站(WWW.)站内文章除注明原创外，均为转载、整理或搜集自网络。欢迎任何形式的转载，转载请注明出处。