Rich Felker
122d67f846
first, the condition (mem && k < p) is redundant, because mem being nonzero implies the needle is periodic with period exactly p, in which case any byte that appears in the needle must appear in the last p bytes of the needle, bounding the shift (k) by p. second, the whole point of replacing the shift k by mem (=l-p) is to prevent shifting by less than mem when discarding the memory on shift, in which case linear time could not be guaranteed. but as written, the check also replaced shifts greater than mem by mem, reducing the benefit of the shift. there is no possible benefit to this reduction of the shift; since mem is being cleared, the full shift is valid and more optimal. so only replace the shift by mem when it would be less than mem.
150 lines
3.4 KiB
C
150 lines
3.4 KiB
C
#define _GNU_SOURCE
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#include <string.h>
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#include <stdint.h>
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static char *twobyte_memmem(const unsigned char *h, size_t k, const unsigned char *n)
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{
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uint16_t nw = n[0]<<8 | n[1], hw = h[0]<<8 | h[1];
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for (h+=2, k-=2; k; k--, hw = hw<<8 | *h++)
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if (hw == nw) return (char *)h-2;
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return hw == nw ? (char *)h-2 : 0;
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}
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static char *threebyte_memmem(const unsigned char *h, size_t k, const unsigned char *n)
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{
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uint32_t nw = n[0]<<24 | n[1]<<16 | n[2]<<8;
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uint32_t hw = h[0]<<24 | h[1]<<16 | h[2]<<8;
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for (h+=3, k-=3; k; k--, hw = (hw|*h++)<<8)
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if (hw == nw) return (char *)h-3;
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return hw == nw ? (char *)h-3 : 0;
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}
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static char *fourbyte_memmem(const unsigned char *h, size_t k, const unsigned char *n)
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{
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uint32_t nw = n[0]<<24 | n[1]<<16 | n[2]<<8 | n[3];
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uint32_t hw = h[0]<<24 | h[1]<<16 | h[2]<<8 | h[3];
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for (h+=4, k-=4; k; k--, hw = hw<<8 | *h++)
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if (hw == nw) return (char *)h-4;
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return hw == nw ? (char *)h-4 : 0;
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}
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#define MAX(a,b) ((a)>(b)?(a):(b))
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#define MIN(a,b) ((a)<(b)?(a):(b))
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#define BITOP(a,b,op) \
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((a)[(size_t)(b)/(8*sizeof *(a))] op (size_t)1<<((size_t)(b)%(8*sizeof *(a))))
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static char *twoway_memmem(const unsigned char *h, const unsigned char *z, const unsigned char *n, size_t l)
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{
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size_t i, ip, jp, k, p, ms, p0, mem, mem0;
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size_t byteset[32 / sizeof(size_t)] = { 0 };
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size_t shift[256];
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/* Computing length of needle and fill shift table */
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for (i=0; i<l; i++)
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BITOP(byteset, n[i], |=), shift[n[i]] = i+1;
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/* Compute maximal suffix */
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ip = -1; jp = 0; k = p = 1;
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while (jp+k<l) {
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if (n[ip+k] == n[jp+k]) {
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if (k == p) {
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jp += p;
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k = 1;
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} else k++;
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} else if (n[ip+k] > n[jp+k]) {
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jp += k;
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k = 1;
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p = jp - ip;
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} else {
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ip = jp++;
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k = p = 1;
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}
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}
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ms = ip;
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p0 = p;
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/* And with the opposite comparison */
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ip = -1; jp = 0; k = p = 1;
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while (jp+k<l) {
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if (n[ip+k] == n[jp+k]) {
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if (k == p) {
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jp += p;
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k = 1;
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} else k++;
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} else if (n[ip+k] < n[jp+k]) {
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jp += k;
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k = 1;
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p = jp - ip;
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} else {
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ip = jp++;
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k = p = 1;
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}
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}
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if (ip+1 > ms+1) ms = ip;
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else p = p0;
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/* Periodic needle? */
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if (memcmp(n, n+p, ms+1)) {
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mem0 = 0;
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p = MAX(ms, l-ms-1) + 1;
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} else mem0 = l-p;
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mem = 0;
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/* Search loop */
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for (;;) {
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/* If remainder of haystack is shorter than needle, done */
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if (z-h < l) return 0;
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/* Check last byte first; advance by shift on mismatch */
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if (BITOP(byteset, h[l-1], &)) {
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k = l-shift[h[l-1]];
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if (k) {
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if (k < mem) k = mem;
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h += k;
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mem = 0;
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continue;
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}
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} else {
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h += l;
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mem = 0;
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continue;
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}
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/* Compare right half */
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for (k=MAX(ms+1,mem); k<l && n[k] == h[k]; k++);
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if (k < l) {
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h += k-ms;
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mem = 0;
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continue;
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}
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/* Compare left half */
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for (k=ms+1; k>mem && n[k-1] == h[k-1]; k--);
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if (k <= mem) return (char *)h;
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h += p;
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mem = mem0;
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}
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}
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void *memmem(const void *h0, size_t k, const void *n0, size_t l)
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{
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const unsigned char *h = h0, *n = n0;
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/* Return immediately on empty needle */
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if (!l) return (void *)h;
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/* Return immediately when needle is longer than haystack */
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if (k<l) return 0;
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/* Use faster algorithms for short needles */
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h = memchr(h0, *n, k);
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if (!h || l==1) return (void *)h;
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k -= h - (const unsigned char *)h0;
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if (k<l) return 0;
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if (l==2) return twobyte_memmem(h, k, n);
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if (l==3) return threebyte_memmem(h, k, n);
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if (l==4) return fourbyte_memmem(h, k, n);
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return twoway_memmem(h, h+k, n, l);
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}
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