/*-

 * Original FreeBSD code Copyright (c) 2005-2014 Rich Felker, et al.

 * Modifications for MVM 32 bit search Copyright (c) 2018 Samantha McVey

 *

 * Permission is hereby granted, free of charge, to any person obtaining

 * a copy of this software and associated documentation files (the

 * "Software"), to deal in the Software without restriction, including

 * without limitation the rights to use, copy, modify, merge, publish,

 * distribute, sublicense, and/or sell copies of the Software, and to

 * permit persons to whom the Software is furnished to do so, subject to

 * the following conditions:

 *

 * The above copyright notice and this permission notice shall be

 * included in all copies or substantial portions of the Software.

 *

 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,

 * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF

 * MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.

 * IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY

 * CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT,

 * TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE

 * SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.

 */

#include <string.h>

#include <stdint.h>

/* memmem_two_uint32 uses 64 bit integer reads extensively. Only use it on 64

 * bit processors since it will likely be slower on 32 bit processors. */

#include "moar.h"

#if 8 <= MVM_PTR_SIZE && defined(MVM_CAN_UNALIGNED_INT64)

#define USE_MEMMEM_TWO_UINT32 1

#endif

/* This is a modification of the memmem used in FreeBSD to allow us to quickly

 * search 32 bit strings. This is much more efficient than searching by byte

 * since we can skip every 4 bytes instead of every 1 byte, as well as the fact

 * that in a 32 bit integer, some of the bytes will be empty, making it even

 * less efficient.

 * The only caveats is the table uses a modulus so it can only jump to the next

 * codepoint of the same modulus. */

/* For memmem_one32 we just look for a single 32 bit integer in the haystack,

 * simple. */

static uint32_t * memmem_one_uint32(const uint32_t *h0, const uint32_t *n0, const uint32_t *end_h0) {

  uint32_t *h           = (uint32_t*)h0;

  const uint32_t  n     = *n0;

  const uint32_t *end_h = end_h0 - 1;

  for (; h <= end_h; h++) {

    if (*h == n) return h;

  }

  return NULL;

}

/* For memmem_two_uint32 we work similarly to memmem_one_uint32 except we read

 * the needle of length 2 as a 64 bit integer, progressing along the haystack

 * 32 bits at a time, and comparing the 64 bit needle to the haystack's pointer

 * casted as a 64 bit integer. */

#if defined(USE_MEMMEM_TWO_UINT32)

static uint32_t * memmem_two_uint32(const uint32_t *h0, const uint32_t *n0, const uint32_t *end_h0) {

  uint32_t *h           = (uint32_t*)h0;

  const uint64_t  n     = *((uint64_t*)n0);

  const uint32_t *end_h = end_h0 - 2;

  for (; h <= end_h; h++) {

    if (*((uint64_t*)h) == n) return h;

  }

  return NULL;

}

#endif

#define MAX(a,b) ((a)>(b)?(a):(b))

#define MIN(a,b) ((a)<(b)?(a):(b))

#define BITOP(a,b,op) \

 ((a)[(size_t)(b)/(8*sizeof *(a))] op (size_t)1<<((size_t)(b)%(8*sizeof *(a))))

/* Original FreeBSD comment:

 * Two Way string search algorithm, with a bad shift table applied to the last

 * byte of the window. A bit array marks which entries in the shift table are

 * initialized to avoid fully initializing a 1kb/2kb table.

 *

 * Reference: CROCHEMORE M., PERRIN D., 1991, Two-way string-matching,

 * Journal of the ACM 38(3):651-675

 */

/* The modulus number is the length of the shift table. We use this to quickly

 * lookup a codepoint and see how much we can shift forward. Caveat: if two numbers

 * have the same modulus we can only shift as much forward as much as a codepoint

 * with that modulus. */

#define MODNUMBER 256

#define MOD_OP(a) ((a) % MODNUMBER)

static char *twoway_memmem_uint32(const uint32_t *h, const uint32_t *z, const uint32_t *n, size_t l)

{

  size_t i, ip, jp, k, p, ms, p0, mem, mem0;

  size_t byteset[32 / sizeof(size_t)] = { 0 };

  uint32_t shift[MODNUMBER];

  /* Computing length of needle and fill shift table */

  for (i=0; i<l; i++) {

    BITOP(byteset, MOD_OP(n[i]), |=);

    shift[MOD_OP(n[i])] = i+1;

  }

  /* Compute maximal suffix */

  ip = -1; jp = 0; k = p = 1;

  while (jp+k<l) {

    if (n[ip+k] == n[jp+k]) {

      if (k == p) {

        jp += p;

        k = 1;

      } else k++;

    } else if (n[ip+k] > n[jp+k]) {

      jp += k;

      k = 1;

      p = jp - ip;

    } else {

      ip = jp++;

      k = p = 1;

    }

  }

  ms = ip;

  p0 = p;

  /* And with the opposite comparison */

  ip = -1; jp = 0; k = p = 1;

  while (jp+k<l) {

    if (n[ip+k] == n[jp+k]) {

      if (k == p) {

        jp += p;

        k = 1;

      } else k++;

    } else if (n[ip+k] < n[jp+k]) {

      jp += k;

      k = 1;

      p = jp - ip;

    } else {

      ip = jp++;

      k = p = 1;

    }

  }

  if (ip+1 > ms+1) ms = ip;

  else p = p0;

  /* Periodic needle? */

  if (memcmp(n, n+p, ms+1)) {

    mem0 = 0;

    p = MAX(ms, l-ms-1) + 1;

  } else mem0 = l-p;

  mem = 0;

  /* Search loop */

  for (;;) {

    /* If remainder of haystack is shorter than needle, done */

    if (z-h < l) {

      return 0;

    }

    /* Check last byte first; advance by shift on mismatch */

    if (BITOP(byteset, MOD_OP(h[l-1]), &)) {

      k = l-shift[MOD_OP(h[l-1])];

      if (k) {

        if (mem0 && mem && k < p) k = l-p;

        h += k;

        mem = 0;

        continue;

      }

    } else {

      h += l;

      mem = 0;

      continue;

    }

    /* Compare right half */

    for (k=MAX(ms+1,mem); k<l && n[k] == h[k]; k++);

    if (k < l) {

      h += k-ms;

      mem = 0;

      continue;

    }

    /* Compare left half */

    for (k=ms+1; k>mem && n[k-1] == h[k-1]; k--);

    if (k <= mem) {

      return (char *)h;

    }

    h += p;

    mem = mem0;

  }

}

/* Finds the memory location of the needle in the haystack. Arguments are the

 * memory location of the start of the Haystack, the memory location of the start

 * of the needle and the needle length as well as the Haystack length.

 * H_len and n_len are measured the size of a uint32_t. */

void * memmem_uint32(const void *h0, size_t H_len, const void *n0, size_t n_len)

{

  const uint32_t *h = (uint32_t*)h0,

                 *n = (uint32_t*)n0;

  /* The empty string can be found at the start of any string. */

  if (!n_len) return (void *)h;

  /* Return immediately when needle is longer than haystack. */

  if (H_len < n_len)

    return NULL;

#if defined(USE_MEMMEM_TWO_UINT32)

  if (n_len == 1) {

    return memmem_one_uint32(h, n, h+H_len);

  }

  /* Otherwise do a search for the first two uint32_t integers at the start

   * of the needle. */

  else {

    h = memmem_two_uint32(h, n, h+H_len);

  }

  /* If nothing found or if we have a needle of length 2, we already have

   * our result. */

  if (!h || n_len == 2)

    return (void *)h;

#else

  /* With needle length 1 use fast search for only one uint32_t. */

  h = memmem_one_uint32(h, n, h+H_len);

  if (!h || n_len == 1)

    return (void *)h;

#endif

  /* Since our Haystack (may) have been moved forward to the first match

   * of the start of the needle, also reduce the Haystack's length to match. */

  H_len -= h - (uint32_t*)h0;

  /* No more work to do if the needle is longer than where we are now. */

  if (H_len < n_len)

    return NULL;

  return twoway_memmem_uint32(h, h+H_len, n, n_len);

}