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/*#************************************************************************#*/
/*#-------------------------------------------------------------------------*/
/*# */
/*# FUNCTION NAME: memset() */
/*# */
/*# PARAMETERS: void* dst; Destination address. */
/*# int c; Value of byte to write. */
/*# int len; Number of bytes to write. */
/*# */
/*# RETURNS: dst. */
/*# */
/*# DESCRIPTION: Sets the memory dst of length len bytes to c, as standard. */
/*# Framework taken from memcpy. This routine is */
/*# very sensitive to compiler changes in register allocation. */
/*# Should really be rewritten to avoid this problem. */
/*# */
/*#-------------------------------------------------------------------------*/
/*# */
/*# HISTORY */
/*# */
/*# DATE NAME CHANGES */
/*# ---- ---- ------- */
/*# 990713 HP Tired of watching this function (or */
/*# really, the nonoptimized generic */
/*# implementation) take up 90% of simulator */
/*# output. Measurements needed. */
/*# */
/*#-------------------------------------------------------------------------*/
#include <linux/types.h>
/* No, there's no macro saying 12*4, since it is "hard" to get it into
the asm in a good way. Thus better to expose the problem everywhere.
*/
/* Assuming 1 cycle per dword written or read (ok, not really true), and
one per instruction, then 43+3*(n/48-1) <= 24+24*(n/48-1)
so n >= 45.7; n >= 0.9; we win on the first full 48-byte block to set. */
#define ZERO_BLOCK_SIZE (1*12*4)
void *memset(void *pdst,
int c,
size_t plen)
{
/* Ok. Now we want the parameters put in special registers.
Make sure the compiler is able to make something useful of this. */
register char *return_dst __asm__ ("r10") = pdst;
register int n __asm__ ("r12") = plen;
register int lc __asm__ ("r11") = c;
/* Most apps use memset sanely. Only those memsetting about 3..4
bytes or less get penalized compared to the generic implementation
- and that's not really sane use. */
/* Ugh. This is fragile at best. Check with newer GCC releases, if
they compile cascaded "x |= x << 8" sanely! */
__asm__("movu.b %0,$r13\n\t"
"lslq 8,$r13\n\t"
"move.b %0,$r13\n\t"
"move.d $r13,%0\n\t"
"lslq 16,$r13\n\t"
"or.d $r13,%0"
: "=r" (lc) : "0" (lc) : "r13");
{
register char *dst __asm__ ("r13") = pdst;
/* This is NONPORTABLE, but since this whole routine is */
/* grossly nonportable that doesn't matter. */
if (((unsigned long) pdst & 3) != 0
/* Oops! n=0 must be a legal call, regardless of alignment. */
&& n >= 3)
{
if ((unsigned long)dst & 1)
{
*dst = (char) lc;
n--;
dst++;
}
if ((unsigned long)dst & 2)
{
*(short *)dst = lc;
n -= 2;
dst += 2;
}
}
/* Now the fun part. For the threshold value of this, check the equation
above. */
/* Decide which copying method to use. */
if (n >= ZERO_BLOCK_SIZE)
{
/* For large copies we use 'movem' */
/* It is not optimal to tell the compiler about clobbering any
registers; that will move the saving/restoring of those registers
to the function prologue/epilogue, and make non-movem sizes
suboptimal.
This method is not foolproof; it assumes that the "asm reg"
declarations at the beginning of the function really are used
here (beware: they may be moved to temporary registers).
This way, we do not have to save/move the registers around into
temporaries; we can safely use them straight away.
If you want to check that the allocation was right; then
check the equalities in the first comment. It should say
"r13=r13, r12=r12, r11=r11" */
__asm__ volatile ("\n\
;; Check that the following is true (same register names on \n\
;; both sides of equal sign, as in r8=r8): \n\
;; %0=r13, %1=r12, %4=r11 \n\
;; \n\
;; Save the registers we'll clobber in the movem process \n\
;; on the stack. Don't mention them to gcc, it will only be \n\
;; upset. \n\
subq 11*4,$sp \n\
movem $r10,[$sp] \n\
\n\
move.d $r11,$r0 \n\
move.d $r11,$r1 \n\
move.d $r11,$r2 \n\
move.d $r11,$r3 \n\
move.d $r11,$r4 \n\
move.d $r11,$r5 \n\
move.d $r11,$r6 \n\
move.d $r11,$r7 \n\
move.d $r11,$r8 \n\
move.d $r11,$r9 \n\
move.d $r11,$r10 \n\
\n\
;; Now we've got this: \n\
;; r13 - dst \n\
;; r12 - n \n\
\n\
;; Update n for the first loop \n\
subq 12*4,$r12 \n\
0: \n\
subq 12*4,$r12 \n\
bge 0b \n\
movem $r11,[$r13+] \n\
\n\
addq 12*4,$r12 ;; compensate for last loop underflowing n \n\
\n\
;; Restore registers from stack \n\
movem [$sp+],$r10"
/* Outputs */ : "=r" (dst), "=r" (n)
/* Inputs */ : "0" (dst), "1" (n), "r" (lc));
}
/* 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 (<44 bytes) is copied here.
This will work since 'movem' will have updated src,dst,n. */
while ( n >= 16 )
{
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
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) lc;
break;
case 2:
*(short*)dst = (short) lc;
break;
case 3:
*((short*)dst)++ = (short) lc;
*(char*)dst = (char) lc;
break;
case 4:
*((long*)dst)++ = lc;
break;
case 5:
*((long*)dst)++ = lc;
*(char*)dst = (char) lc;
break;
case 6:
*((long*)dst)++ = lc;
*(short*)dst = (short) lc;
break;
case 7:
*((long*)dst)++ = lc;
*((short*)dst)++ = (short) lc;
*(char*)dst = (char) lc;
break;
case 8:
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
break;
case 9:
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*(char*)dst = (char) lc;
break;
case 10:
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*(short*)dst = (short) lc;
break;
case 11:
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*((short*)dst)++ = (short) lc;
*(char*)dst = (char) lc;
break;
case 12:
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
break;
case 13:
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*(char*)dst = (char) lc;
break;
case 14:
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*(short*)dst = (short) lc;
break;
case 15:
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*((long*)dst)++ = lc;
*((short*)dst)++ = (short) lc;
*(char*)dst = (char) lc;
break;
}
}
return return_dst; /* destination pointer. */
} /* memset() */
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