-/*
- * Context of comparison operation.
- */
-struct context
-{
- /*
- * Data on one input string being compared.
- */
- struct string_data
- {
- /* The string to be compared. */
- const char *data;
-
- /* The length of the string to be compared. */
- int data_length;
-
- /* The number of characters inserted or deleted. */
- int edit_count;
- }
- string[2];
-
- #ifdef MINUS_H_FLAG
-
- /* This corresponds to the diff -H flag. With this heuristic, for
- strings with a constant small density of changes, the algorithm is
- linear in the strings size. This is unlikely in typical uses of
- fstrcmp, and so is usually compiled out. Besides, there is no
- interface to set it true. */
- int heuristic;
-
- #endif
-
- /* Vector, indexed by diagonal, containing 1 + the X coordinate of the
- point furthest along the given diagonal in the forward search of the
- edit matrix. */
- int *fdiag;
-
- /* Vector, indexed by diagonal, containing the X coordinate of the point
- furthest along the given diagonal in the backward search of the edit
- matrix. */
- int *bdiag;
-
- /* Edit scripts longer than this are too expensive to compute. */
- int too_expensive;
-
- /* Snakes bigger than this are considered `big'. */
- #define SNAKE_LIMIT 20
-};
-
-struct partition
-{
- /* Midpoints of this partition. */
- int xmid, ymid;
-
- /* Nonzero if low half will be analyzed minimally. */
- int lo_minimal;
-
- /* Likewise for high half. */
- int hi_minimal;
-};
-
-
-/* NAME
- diag - find diagonal path
-
- SYNOPSIS
- int diag(int xoff, int xlim, int yoff, int ylim, int minimal,
- struct partition *part, struct context *ctxt);
-
- DESCRIPTION
- Find the midpoint of the shortest edit script for a specified
- portion of the two strings.
-
- Scan from the beginnings of the strings, and simultaneously from
- the ends, doing a breadth-first search through the space of
- edit-sequence. When the two searches meet, we have found the
- midpoint of the shortest edit sequence.
-
- If MINIMAL is nonzero, find the minimal edit script regardless
- of expense. Otherwise, if the search is too expensive, use
- heuristics to stop the search and report a suboptimal answer.
-
- RETURNS
- Set PART->(XMID,YMID) to the midpoint (XMID,YMID). The diagonal
- number XMID - YMID equals the number of inserted characters
- minus the number of deleted characters (counting only characters
- before the midpoint). Return the approximate edit cost; this is
- the total number of characters inserted or deleted (counting
- only characters before the midpoint), unless a heuristic is used
- to terminate the search prematurely.
-
- Set PART->LEFT_MINIMAL to nonzero iff the minimal edit script
- for the left half of the partition is known; similarly for
- PART->RIGHT_MINIMAL.
-
- CAVEAT
- This function assumes that the first characters of the specified
- portions of the two strings do not match, and likewise that the
- last characters do not match. The caller must trim matching
- characters from the beginning and end of the portions it is
- going to specify.
-
- If we return the "wrong" partitions, the worst this can do is
- cause suboptimal diff output. It cannot cause incorrect diff
- output. */
-
-static int
-diag (int xoff, int xlim, int yoff, int ylim, int minimal,
- struct partition *part, struct context *ctxt)
-{
- int *const fd = ctxt->fdiag; /* Give the compiler a chance. */
- int *const bd = ctxt->bdiag; /* Additional help for the compiler. */
- const char *const xv = ctxt->string[0].data; /* Still more help for the compiler. */
- const char *const yv = ctxt->string[1].data; /* And more and more . . . */
- const int dmin = xoff - ylim; /* Minimum valid diagonal. */
- const int dmax = xlim - yoff; /* Maximum valid diagonal. */
- const int fmid = xoff - yoff; /* Center diagonal of top-down search. */
- const int bmid = xlim - ylim; /* Center diagonal of bottom-up search. */
- int fmin = fmid;
- int fmax = fmid; /* Limits of top-down search. */
- int bmin = bmid;
- int bmax = bmid; /* Limits of bottom-up search. */
- int c; /* Cost. */
- int odd = (fmid - bmid) & 1;
-
- /*
- * True if southeast corner is on an odd diagonal with respect
- * to the northwest.
- */
- fd[fmid] = xoff;
- bd[bmid] = xlim;
- for (c = 1;; ++c)
- {
- int d; /* Active diagonal. */
- int big_snake;
-
- big_snake = 0;
- /* Extend the top-down search by an edit step in each diagonal. */
- if (fmin > dmin)
- fd[--fmin - 1] = -1;
- else
- ++fmin;
- if (fmax < dmax)
- fd[++fmax + 1] = -1;
- else
- --fmax;
- for (d = fmax; d >= fmin; d -= 2)
- {
- int x;
- int y;
- int oldx;
- int tlo;
- int thi;
-
- tlo = fd[d - 1],
- thi = fd[d + 1];
-
- if (tlo >= thi)
- x = tlo + 1;
- else
- x = thi;
- oldx = x;
- y = x - d;
- while (x < xlim && y < ylim && xv[x] == yv[y])
- {
- ++x;
- ++y;
- }
- if (x - oldx > SNAKE_LIMIT)
- big_snake = 1;
- fd[d] = x;
- if (odd && bmin <= d && d <= bmax && bd[d] <= x)
- {
- part->xmid = x;
- part->ymid = y;
- part->lo_minimal = part->hi_minimal = 1;
- return 2 * c - 1;
- }
- }
- /* Similarly extend the bottom-up search. */
- if (bmin > dmin)
- bd[--bmin - 1] = INT_MAX;
- else
- ++bmin;
- if (bmax < dmax)
- bd[++bmax + 1] = INT_MAX;
- else
- --bmax;
- for (d = bmax; d >= bmin; d -= 2)
- {
- int x;
- int y;
- int oldx;
- int tlo;
- int thi;
-
- tlo = bd[d - 1],
- thi = bd[d + 1];
- if (tlo < thi)
- x = tlo;
- else
- x = thi - 1;
- oldx = x;
- y = x - d;
- while (x > xoff && y > yoff && xv[x - 1] == yv[y - 1])
- {
- --x;
- --y;
- }
- if (oldx - x > SNAKE_LIMIT)
- big_snake = 1;
- bd[d] = x;
- if (!odd && fmin <= d && d <= fmax && x <= fd[d])
- {
- part->xmid = x;
- part->ymid = y;
- part->lo_minimal = part->hi_minimal = 1;
- return 2 * c;
- }
- }
-
- if (minimal)
- continue;
-
-#ifdef MINUS_H_FLAG
- /* Heuristic: check occasionally for a diagonal that has made lots
- of progress compared with the edit distance. If we have any
- such, find the one that has made the most progress and return
- it as if it had succeeded.
-
- With this heuristic, for strings with a constant small density
- of changes, the algorithm is linear in the strings size. */
- if (c > 200 && big_snake && ctxt->heuristic)
- {
- int best;
-
- best = 0;
- for (d = fmax; d >= fmin; d -= 2)
- {
- int dd;
- int x;
- int y;
- int v;
-
- dd = d - fmid;
- x = fd[d];
- y = x - d;
- v = (x - xoff) * 2 - dd;
-
- if (v > 12 * (c + (dd < 0 ? -dd : dd)))
- {
- if
- (
- v > best
- &&
- xoff + SNAKE_LIMIT <= x
- &&
- x < xlim
- &&
- yoff + SNAKE_LIMIT <= y
- &&
- y < ylim
- )
- {
- /* We have a good enough best diagonal; now insist
- that it end with a significant snake. */
- int k;
-
- for (k = 1; xv[x - k] == yv[y - k]; k++)
- {
- if (k == SNAKE_LIMIT)
- {
- best = v;
- part->xmid = x;
- part->ymid = y;
- break;
- }
- }
- }
- }
- }
- if (best > 0)
- {
- part->lo_minimal = 1;
- part->hi_minimal = 0;
- return 2 * c - 1;
- }
- best = 0;
- for (d = bmax; d >= bmin; d -= 2)
- {
- int dd;
- int x;
- int y;
- int v;
-
- dd = d - bmid;
- x = bd[d];
- y = x - d;
- v = (xlim - x) * 2 + dd;
-
- if (v > 12 * (c + (dd < 0 ? -dd : dd)))
- {
- if (v > best && xoff < x && x <= xlim - SNAKE_LIMIT &&
- yoff < y && y <= ylim - SNAKE_LIMIT)
- {
- /* We have a good enough best diagonal; now insist
- that it end with a significant snake. */
- int k;
-
- for (k = 0; xv[x + k] == yv[y + k]; k++)
- {
- if (k == SNAKE_LIMIT - 1)
- {
- best = v;
- part->xmid = x;
- part->ymid = y;
- break;
- }
- }
- }
- }
- }
- if (best > 0)
- {
- part->lo_minimal = 0;
- part->hi_minimal = 1;
- return 2 * c - 1;
- }
- }
-#endif /* MINUS_H_FLAG */
-
- /* Heuristic: if we've gone well beyond the call of duty, give up
- and report halfway between our best results so far. */
- if (c >= ctxt->too_expensive)
- {
- int fxybest;
- int fxbest;
- int bxybest;
- int bxbest;
-
- /* Pacify `gcc -Wall'. */
- fxbest = 0;
- bxbest = 0;
-
- /* Find forward diagonal that maximizes X + Y. */
- fxybest = -1;
- for (d = fmax; d >= fmin; d -= 2)
- {
- int x;
- int y;
-
- x = fd[d] < xlim ? fd[d] : xlim;
- y = x - d;
-
- if (ylim < y)
- {
- x = ylim + d;
- y = ylim;
- }
- if (fxybest < x + y)
- {
- fxybest = x + y;
- fxbest = x;
- }
- }
- /* Find backward diagonal that minimizes X + Y. */
- bxybest = INT_MAX;
- for (d = bmax; d >= bmin; d -= 2)
- {
- int x;
- int y;
-
- x = xoff > bd[d] ? xoff : bd[d];
- y = x - d;
-
- if (y < yoff)
- {
- x = yoff + d;
- y = yoff;
- }
- if (x + y < bxybest)
- {
- bxybest = x + y;
- bxbest = x;
- }
- }
- /* Use the better of the two diagonals. */
- if ((xlim + ylim) - bxybest < fxybest - (xoff + yoff))
- {
- part->xmid = fxbest;
- part->ymid = fxybest - fxbest;
- part->lo_minimal = 1;
- part->hi_minimal = 0;
- }
- else
- {
- part->xmid = bxbest;
- part->ymid = bxybest - bxbest;
- part->lo_minimal = 0;
- part->hi_minimal = 1;
- }
- return 2 * c - 1;
- }
- }
-}
-
-
-/* NAME
- compareseq - find edit sequence
-
- SYNOPSIS
- void compareseq(int xoff, int xlim, int yoff, int ylim, int minimal,
- struct context *ctxt);
-
- DESCRIPTION
- Compare in detail contiguous subsequences of the two strings
- which are known, as a whole, to match each other.
-
- The subsequence of string 0 is [XOFF, XLIM) and likewise for
- string 1.
-
- Note that XLIM, YLIM are exclusive bounds. All character
- numbers are origin-0.
-
- If MINIMAL is nonzero, find a minimal difference no matter how
- expensive it is. */
-
-static void
-compareseq (int xoff, int xlim, int yoff, int ylim, int minimal,
- struct context *ctxt)
-{
- const char *const xv = ctxt->string[0].data; /* Help the compiler. */
- const char *const yv = ctxt->string[1].data;
-
- /* Slide down the bottom initial diagonal. */
- while (xoff < xlim && yoff < ylim && xv[xoff] == yv[yoff])
- {
- ++xoff;
- ++yoff;
- }
-
- /* Slide up the top initial diagonal. */
- while (xlim > xoff && ylim > yoff && xv[xlim - 1] == yv[ylim - 1])
- {
- --xlim;
- --ylim;
- }
-
- /* Handle simple cases. */
- if (xoff == xlim)
- {
- while (yoff < ylim)
- {
- ctxt->string[1].edit_count++;
- ++yoff;
- }
- }
- else if (yoff == ylim)
- {
- while (xoff < xlim)
- {
- ctxt->string[0].edit_count++;
- ++xoff;
- }
- }
- else
- {
- int c;
- struct partition part;
-
- /* Find a point of correspondence in the middle of the strings. */
- c = diag (xoff, xlim, yoff, ylim, minimal, &part, ctxt);
- if (c == 1)
- {
-#if 0
- /* This should be impossible, because it implies that one of
- the two subsequences is empty, and that case was handled
- above without calling `diag'. Let's verify that this is
- true. */
- abort ();
-#else
- /* The two subsequences differ by a single insert or delete;
- record it and we are done. */
- if (part.xmid - part.ymid < xoff - yoff)
- ctxt->string[1].edit_count++;
- else
- ctxt->string[0].edit_count++;
-#endif
- }
- else
- {
- /* Use the partitions to split this problem into subproblems. */
- compareseq (xoff, part.xmid, yoff, part.ymid, part.lo_minimal, ctxt);
- compareseq (part.xmid, xlim, part.ymid, ylim, part.hi_minimal, ctxt);
- }
- }
-}
projects / gnulib.git / blobdiff