Hash :
e2eeaaf7
Author :
Date :
2010-11-27T01:29:56
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#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <sys/time.h>
#include "miner.h"
#ifdef WANT_VIA_PADLOCK
#define ___constant_swab32(x) ((uint32_t)( \
(((uint32_t)(x) & (uint32_t)0x000000ffUL) << 24) | \
(((uint32_t)(x) & (uint32_t)0x0000ff00UL) << 8) | \
(((uint32_t)(x) & (uint32_t)0x00ff0000UL) >> 8) | \
(((uint32_t)(x) & (uint32_t)0xff000000UL) >> 24)))
static inline uint32_t swab32(uint32_t v)
{
return ___constant_swab32(v);
}
static void via_sha256(void *hash, void *buf, unsigned len)
{
unsigned stat = 0;
asm volatile(".byte 0xf3, 0x0f, 0xa6, 0xd0"
:"+S"(buf), "+a"(stat)
:"c"(len), "D" (hash)
:"memory");
}
bool scanhash_via(unsigned char *midstate, const unsigned char *data_in,
unsigned char *hash1, unsigned char *hash,
unsigned long *hashes_done)
{
unsigned char data[128] __attribute__((aligned(128)));
unsigned char tmp_hash1[32] __attribute__((aligned(32)));
uint32_t *hash32 = (uint32_t *) hash;
uint32_t *nonce = (uint32_t *)(data + 12);
uint32_t n = 0;
unsigned long stat_ctr = 0;
int i;
/* bitcoin gives us big endian input, but via wants LE,
* so we reverse the swapping bitcoin has already done (extra work)
*/
for (i = 0; i < 128/4; i++) {
uint32_t *data32 = (uint32_t *) data;
data32[i] = swab32(((uint32_t *)data_in)[i]);
}
while (1) {
n++;
*nonce = n;
/* first SHA256 transform */
memcpy(tmp_hash1, sha256_init_state, 32);
via_sha256(tmp_hash1, data, 80); /* or maybe 128? */
for (i = 0; i < 32/4; i++)
((uint32_t *)tmp_hash1)[i] =
swab32(((uint32_t *)tmp_hash1)[i]);
/* second SHA256 transform */
memcpy(hash, sha256_init_state, 32);
via_sha256(hash, tmp_hash1, 32);
stat_ctr++;
if (hash32[7] == 0) {
char *hexstr;
hexstr = bin2hex(hash, 32);
fprintf(stderr,
"DBG: found zeroes in hash:\n%s\n",
hexstr);
free(hexstr);
*hashes_done = stat_ctr;
return true;
}
if ((n & 0xffffff) == 0) {
if (opt_debug)
fprintf(stderr, "DBG: end of nonce range\n");
*hashes_done = stat_ctr;
return false;
}
}
}
#endif /* WANT_VIA_PADLOCK */