x86 – nullteilerfrei

After a couple of tests, it turns out that the very simple

#include <limits.h>

inline void memzap(void *dest, unsigned long count) {
    asm( "cld"
#   if ULONG_MAX == 0xffffffff
    "\n" "andl $3, %%ecx"
    "\n" "rep  stosb"
    "\n" "movl %%ebx, %%ecx"
    "\n" "shrl $2, %%ecx"
    "\n" "rep  stosl"
#   else
    "\n" "andq $7, %%rcx"
    "\n" "rep  stosb"
    "\n" "movq %%rbx, %%rcx"
    "\n" "shrq $3, %%rcx"
    "\n" "rep  stosq"
#   endif
      : "=c" (count), "=D" (dest), "=b" (count)
      :  "c" (count),  "D" (dest),  "b" (count), "a" (0)
    );
}

is the fastest way to zero out a large block of memory, which is not very surprising. It is about 4 to 5 times faster than memset and about as fast as new [], if I can trust @tobi on that matter. I tried using MMX registers, but anything that involves actually looping over the memory region will be about as fast as memset. The only way to get a bit of speed is using the rep opcode. Tiny Edit: The above code is much more safe to compile on both 64 and 32 bit computers.

For reasons I have not yet been able to figure out, @tobi is making me implement a couple of very rudimentary routines in x86 GCC inline assembler because he wants them faster than possible for mere mortal C. The first was a routine to calculate $\lfloor\log_2(n)\rfloor$ for $n\in\mathbb{N}$ and the second one was to zero out a large block of memory. For instance,

unsigned inline log2int(unsigned x) {
    unsigned l;
    asm("bsrl %1, %0" : "=r" (l) : "r" (x));
    return ( 1 << l == x ) ? l : l + 1;
}

is about 50 times faster than the C-native Version

unsigned inline log2int(unsigned x) {
   unsigned l = 0;
   while(x > (1<<l)) l++;
   return l;
}

even after optimization. For some reason, I found it tricky to google up the official intel x86 opcode reference ((Opcode Reference Part 1)) ((Opcode Reference Part 2)), so I am linking these here.

Zero out large blocks of memory

Official Intel ASM Opcode Reference