if ((op0->architecture == v9) != (op1->architecture == v9)) return (op0->architecture == v9) - (op1->architecture == v9);
compare_opcodes (a, b)     char *a, *b;{  struct sparc_opcode *op0 = (struct sparc_opcode *) a;  struct sparc_opcode *op1 = (struct sparc_opcode *) b;  unsigned long int match0 = op0->match, match1 = op1->match;  unsigned long int lose0 = op0->lose, lose1 = op1->lose;  register unsigned int i;  /* If a bit is set in both match and lose, there is something     wrong with the opcode table.  */  if (match0 & lose0)    {      fprintf (stderr, "Internal error:  bad sparc-opcode.h: \"%s\", %#.8lx, %#.8lx\n",	       op0->name, match0, lose0);      op0->lose &= ~op0->match;      lose0 = op0->lose;    }  if (match1 & lose1)    {      fprintf (stderr, "Internal error: bad sparc-opcode.h: \"%s\", %#.8lx, %#.8lx\n",	       op1->name, match1, lose1);      op1->lose &= ~op1->match;      lose1 = op1->lose;    }  /* Because the bits that are variable in one opcode are constant in     another, it is important to order the opcodes in the right order.  */  for (i = 0; i < 32; ++i)    {      unsigned long int x = 1 << i;      int x0 = (match0 & x) != 0;      int x1 = (match1 & x) != 0;      if (x0 != x1)	return x1 - x0;    }  for (i = 0; i < 32; ++i)    {      unsigned long int x = 1 << i;      int x0 = (lose0 & x) != 0;      int x1 = (lose1 & x) != 0;      if (x0 != x1)	return x1 - x0;    }  /* Put non-sparc64 insns ahead of sparc64 ones.  */  if ((op0->architecture == v9) != (op1->architecture == v9))    return (op0->architecture == v9) - (op1->architecture == v9);  /* They are functionally equal.  So as long as the opcode table is     valid, we can put whichever one first we want, on aesthetic grounds.  */  /* Our first aesthetic ground is that aliases defer to real insns.  */  {    int alias_diff = (op0->flags & F_ALIAS) - (op1->flags & F_ALIAS);    if (alias_diff != 0)      /* Put the one that isn't an alias first.  */      return alias_diff;  }  /* Except for aliases, two "identical" instructions had     better have the same opcode.  This is a sanity check on the table.  */  i = strcmp (op0->name, op1->name);  if (i)      if (op0->flags & F_ALIAS) /* If they're both aliases, be arbitrary. */	  return i;      else	  fprintf (stderr,		   "Internal error: bad sparc-opcode.h: \"%s\" == \"%s\"\n",		   op0->name, op1->name);  /* Fewer arguments are preferred.  */  {    int length_diff = strlen (op0->args) - strlen (op1->args);    if (length_diff != 0)      /* Put the one with fewer arguments first.  */      return length_diff;  }  /* Put 1+i before i+1.  */  {    char *p0 = (char *) strchr(op0->args, '+');    char *p1 = (char *) strchr(op1->args, '+');    if (p0 && p1)      {	/* There is a plus in both operands.  Note that a plus	   sign cannot be the first character in args,	   so the following [-1]'s are valid.  */	if (p0[-1] == 'i' && p1[1] == 'i')	  /* op0 is i+1 and op1 is 1+i, so op1 goes first.  */	  return 1;	if (p0[1] == 'i' && p1[-1] == 'i')	  /* op0 is 1+i and op1 is i+1, so op0 goes first.  */	  return -1;      }  }  /* Put 1,i before i,1.  */  {    int i0 = strncmp (op0->args, "i,1", 3) == 0;    int i1 = strncmp (op1->args, "i,1", 3) == 0;    if (i0 ^ i1)      return i0 - i1;  }  /* They are, as far as we can tell, identical.     Since qsort may have rearranged the table partially, there is     no way to tell which one was first in the opcode table as     written, so just say there are equal.  */  return 0;}