io.c

这是一个同样来自贝尔实验室的和UNIX有着渊源的操作系统, 其简洁的设计和实现易于我们学习和理解

					     * sizeof (*linbuf))
		   - linbuf_base;
	}
      linbuf[line] = (char const *) p;

      if ((char const *) p == bufend)
	break;

      if (context <= i && no_diff_means_no_output)
	break;

      line++;

      while (*p++ != '\n')
	;
    }

  /* Done with cache in local variables.  */
  current->linbuf = linbuf;
  current->valid_lines = line;
  current->alloc_lines = alloc_lines;
  current->equivs = cureqs;
  equivs = eqs;
  equivs_alloc = eqs_alloc;
  equivs_index = eqs_index;
}

/* Prepare the end of the text.  Make sure it's initialized.
   Make sure text ends in a newline,
   but remember that we had to add one.  */

static void
prepare_text_end (current)
     struct file_data *current;
{
  size_t buffered_chars = current->buffered_chars;
  char *p = current->buffer;

  if (buffered_chars == 0 || p[buffered_chars - 1] == '\n')
    current->missing_newline = 0;
  else
    {
      p[buffered_chars++] = '\n';
      current->buffered_chars = buffered_chars;
      current->missing_newline = 1;
    }

  /* Don't use uninitialized storage when planting or using sentinels.  */
  if (p)
    bzero (p + buffered_chars, sizeof (word));
}

/* Given a vector of two file_data objects, find the identical
   prefixes and suffixes of each object. */

static void
find_identical_ends (filevec)
     struct file_data filevec[];
{
  word *w0, *w1;
  char *p0, *p1, *buffer0, *buffer1;
  char const *end0, *beg0;
  char const **linbuf0, **linbuf1;
  int i, lines;
  size_t n0, n1, tem;
  int alloc_lines0, alloc_lines1;
  int buffered_prefix, prefix_count, prefix_mask;

  slurp (&filevec[0]);
  if (filevec[0].desc != filevec[1].desc)
    slurp (&filevec[1]);
  else
    {
      filevec[1].buffer = filevec[0].buffer;
      filevec[1].bufsize = filevec[0].bufsize;
      filevec[1].buffered_chars = filevec[0].buffered_chars;
    }
  for (i = 0; i < 2; i++)
    prepare_text_end (&filevec[i]);

  /* Find identical prefix.  */

  p0 = buffer0 = filevec[0].buffer;
  p1 = buffer1 = filevec[1].buffer;

  n0 = filevec[0].buffered_chars;
  n1 = filevec[1].buffered_chars;

  if (p0 == p1)
    /* The buffers are the same; sentinels won't work.  */
    p0 = p1 += n1;
  else
    {
      /* Insert end sentinels, in this case characters that are guaranteed
	 to make the equality test false, and thus terminate the loop.  */

      if (n0 < n1)
	p0[n0] = ~p1[n0];
      else
	p1[n1] = ~p0[n1];

      /* Loop until first mismatch, or to the sentinel characters.  */

      /* Compare a word at a time for speed.  */
      w0 = (word *) p0;
      w1 = (word *) p1;
      while (*w0++ == *w1++)
	;
      --w0, --w1;

      /* Do the last few bytes of comparison a byte at a time.  */
      p0 = (char *) w0;
      p1 = (char *) w1;
      while (*p0++ == *p1++)
	;
      --p0, --p1;

      /* Don't mistakenly count missing newline as part of prefix. */
      if (ROBUST_OUTPUT_STYLE (output_style)
	  && (buffer0 + n0 - filevec[0].missing_newline < p0)
	     !=
	     (buffer1 + n1 - filevec[1].missing_newline < p1))
	--p0, --p1;
    }

  /* Now P0 and P1 point at the first nonmatching characters.  */

  /* Skip back to last line-beginning in the prefix,
     and then discard up to HORIZON_LINES lines from the prefix.  */
  i = horizon_lines;
  while (p0 != buffer0 && (p0[-1] != '\n' || i--))
    --p0, --p1;

  /* Record the prefix.  */
  filevec[0].prefix_end = p0;
  filevec[1].prefix_end = p1;

  /* Find identical suffix.  */

  /* P0 and P1 point beyond the last chars not yet compared.  */
  p0 = buffer0 + n0;
  p1 = buffer1 + n1;

  if (! ROBUST_OUTPUT_STYLE (output_style)
      || filevec[0].missing_newline == filevec[1].missing_newline)
    {
      end0 = p0;	/* Addr of last char in file 0.  */

      /* Get value of P0 at which we should stop scanning backward:
	 this is when either P0 or P1 points just past the last char
	 of the identical prefix.  */
      beg0 = filevec[0].prefix_end + (n0 < n1 ? 0 : n0 - n1);

      /* Scan back until chars don't match or we reach that point.  */
      while (p0 != beg0)
	if (*--p0 != *--p1)
	  {
	    /* Point at the first char of the matching suffix.  */
	    ++p0, ++p1;
	    beg0 = p0;
	    break;
	  }

      /* Are we at a line-beginning in both files?  If not, add the rest of
	 this line to the main body.  Discard up to HORIZON_LINES lines from
	 the identical suffix.  Also, discard one extra line,
	 because shift_boundaries may need it.  */
      i = horizon_lines + !((buffer0 == p0 || p0[-1] == '\n')
			    &&
			    (buffer1 == p1 || p1[-1] == '\n'));
      while (i-- && p0 != end0)
	while (*p0++ != '\n')
	  ;

      p1 += p0 - beg0;
    }

  /* Record the suffix.  */
  filevec[0].suffix_begin = p0;
  filevec[1].suffix_begin = p1;

  /* Calculate number of lines of prefix to save.

     prefix_count == 0 means save the whole prefix;
     we need this with for options like -D that output the whole file.
     We also need it for options like -F that output some preceding line;
     at least we will need to find the last few lines,
     but since we don't know how many, it's easiest to find them all.

     Otherwise, prefix_count != 0.  Save just prefix_count lines at start
     of the line buffer; they'll be moved to the proper location later.
     Handle 1 more line than the context says (because we count 1 too many),
     rounded up to the next power of 2 to speed index computation.  */

  if (no_diff_means_no_output && ! function_regexp_list)
    {
      for (prefix_count = 1;  prefix_count < context + 1;  prefix_count *= 2)
	;
      prefix_mask = prefix_count - 1;
      alloc_lines0
	= prefix_count
	  + GUESS_LINES (0, 0, p0 - filevec[0].prefix_end)
	  + context;
    }
  else
    {
      prefix_count = 0;
      prefix_mask = ~0;
      alloc_lines0 = GUESS_LINES (0, 0, n0);
    }

  lines = 0;
  linbuf0 = (char const **) xmalloc (alloc_lines0 * sizeof (*linbuf0));

  /* If the prefix is needed, find the prefix lines.  */
  if (! (no_diff_means_no_output
	 && filevec[0].prefix_end == p0
	 && filevec[1].prefix_end == p1))
    {
      p0 = buffer0;
      end0 = filevec[0].prefix_end;
      while (p0 != end0)
	{
	  int l = lines++ & prefix_mask;
	  if (l == alloc_lines0)
	    linbuf0 = (char const **) xrealloc (linbuf0, (alloc_lines0 *= 2)
							 * sizeof(*linbuf0));
	  linbuf0[l] = p0;
	  while (*p0++ != '\n')
	    ;
	}
    }
  buffered_prefix = prefix_count && context < lines ? context : lines;

  /* Allocate line buffer 1.  */
  tem = prefix_count ? filevec[1].suffix_begin - buffer1 : n1;

  alloc_lines1
    = (buffered_prefix
       + GUESS_LINES (lines, filevec[1].prefix_end - buffer1, tem)
       + context);
  linbuf1 = (char const **) xmalloc (alloc_lines1 * sizeof (*linbuf1));

  if (buffered_prefix != lines)
    {
      /* Rotate prefix lines to proper location.  */
      for (i = 0;  i < buffered_prefix;  i++)
	linbuf1[i] = linbuf0[(lines - context + i) & prefix_mask];
      for (i = 0;  i < buffered_prefix;  i++)
	linbuf0[i] = linbuf1[i];
    }

  /* Initialize line buffer 1 from line buffer 0.  */
  for (i = 0; i < buffered_prefix; i++)
    linbuf1[i] = linbuf0[i] - buffer0 + buffer1;

  /* Record the line buffer, adjusted so that
     linbuf*[0] points at the first differing line.  */
  filevec[0].linbuf = linbuf0 + buffered_prefix;
  filevec[1].linbuf = linbuf1 + buffered_prefix;
  filevec[0].linbuf_base = filevec[1].linbuf_base = - buffered_prefix;
  filevec[0].alloc_lines = alloc_lines0 - buffered_prefix;
  filevec[1].alloc_lines = alloc_lines1 - buffered_prefix;
  filevec[0].prefix_lines = filevec[1].prefix_lines = lines;
}

/* Largest primes less than some power of two, for nbuckets.  Values range
   from useful to preposterous.  If one of these numbers isn't prime
   after all, don't blame it on me, blame it on primes (6) . . . */
static int const primes[] =
{
  509,
  1021,
  2039,
  4093,
  8191,
  16381,
  32749,
#if 32767 < INT_MAX
  65521,
  131071,
  262139,
  524287,
  1048573,
  2097143,
  4194301,
  8388593,
  16777213,
  33554393,
  67108859,			/* Preposterously large . . . */
  134217689,
  268435399,
  536870909,
  1073741789,
  2147483647,
#endif
  0
};

/* Given a vector of two file_data objects, read the file associated
   with each one, and build the table of equivalence classes.
   Return 1 if either file appears to be a binary file.
   If PRETEND_BINARY is nonzero, pretend they are binary regardless.  */

int
read_files (filevec, pretend_binary)
     struct file_data filevec[];
     int pretend_binary;
{
  int i;
  int skip_test = always_text_flag | pretend_binary;
  int appears_binary = pretend_binary | sip (&filevec[0], skip_test);

  if (filevec[0].desc != filevec[1].desc)
    appears_binary |= sip (&filevec[1], skip_test | appears_binary);
  else
    {
      filevec[1].buffer = filevec[0].buffer;
      filevec[1].bufsize = filevec[0].bufsize;
      filevec[1].buffered_chars = filevec[0].buffered_chars;
    }
  if (appears_binary)
    {
#if HAVE_SETMODE
      setmode (filevec[0].desc, O_BINARY);
      setmode (filevec[1].desc, O_BINARY);
#endif
      return 1;
    }

  find_identical_ends (filevec);

  equivs_alloc = filevec[0].alloc_lines + filevec[1].alloc_lines + 1;
  equivs = (struct equivclass *) xmalloc (equivs_alloc * sizeof (struct equivclass));
  /* Equivalence class 0 is permanently safe for lines that were not
     hashed.  Real equivalence classes start at 1. */
  equivs_index = 1;

  for (i = 0;  primes[i] < equivs_alloc / 3;  i++)
    if (! primes[i])
      abort ();
  nbuckets = primes[i];

  buckets = (int *) xmalloc ((nbuckets + 1) * sizeof (*buckets));
  bzero (buckets++, (nbuckets + 1) * sizeof (*buckets));

  for (i = 0; i < 2; i++)
    find_and_hash_each_line (&filevec[i]);

  filevec[0].equiv_max = filevec[1].equiv_max = equivs_index;

  free (equivs);
  free (buckets - 1);

  return 0;
}