tree.c

GCC编译器源代码

  inline_obstacks = p->inline_obstacks;
}

/* Start allocating on the temporary (per function) obstack.
   This is done in start_function before parsing the function body,
   and before each initialization at top level, and to go back
   to temporary allocation after doing permanent_allocation.  */

void
temporary_allocation ()
{
  /* Note that function_obstack at top level points to temporary_obstack.
     But within a nested function context, it is a separate obstack.  */
  current_obstack = function_obstack;
  expression_obstack = function_obstack;
  rtl_obstack = saveable_obstack = function_maybepermanent_obstack;
  momentary_stack = 0;
  inline_obstacks = 0;
}

/* Start allocating on the permanent obstack but don't
   free the temporary data.  After calling this, call
   `permanent_allocation' to fully resume permanent allocation status.  */

void
end_temporary_allocation ()
{
  current_obstack = &permanent_obstack;
  expression_obstack = &permanent_obstack;
  rtl_obstack = saveable_obstack = &permanent_obstack;
}

/* Resume allocating on the temporary obstack, undoing
   effects of `end_temporary_allocation'.  */

void
resume_temporary_allocation ()
{
  current_obstack = function_obstack;
  expression_obstack = function_obstack;
  rtl_obstack = saveable_obstack = function_maybepermanent_obstack;
}

/* While doing temporary allocation, switch to allocating in such a
   way as to save all nodes if the function is inlined.  Call
   resume_temporary_allocation to go back to ordinary temporary
   allocation.  */

void
saveable_allocation ()
{
  /* Note that function_obstack at top level points to temporary_obstack.
     But within a nested function context, it is a separate obstack.  */
  expression_obstack = current_obstack = saveable_obstack;
}

/* Switch to current obstack CURRENT and maybepermanent obstack SAVEABLE,
   recording the previously current obstacks on a stack.
   This does not free any storage in any obstack.  */

void
push_obstacks (current, saveable)
     struct obstack *current, *saveable;
{
  struct obstack_stack *p
    = (struct obstack_stack *) obstack_alloc (&obstack_stack_obstack,
					      (sizeof (struct obstack_stack)));

  p->current = current_obstack;
  p->saveable = saveable_obstack;
  p->expression = expression_obstack;
  p->rtl = rtl_obstack;
  p->next = obstack_stack;
  obstack_stack = p;

  current_obstack = current;
  expression_obstack = current;
  rtl_obstack = saveable_obstack = saveable;
}

/* Save the current set of obstacks, but don't change them.  */

void
push_obstacks_nochange ()
{
  struct obstack_stack *p
    = (struct obstack_stack *) obstack_alloc (&obstack_stack_obstack,
					      (sizeof (struct obstack_stack)));

  p->current = current_obstack;
  p->saveable = saveable_obstack;
  p->expression = expression_obstack;
  p->rtl = rtl_obstack;
  p->next = obstack_stack;
  obstack_stack = p;
}

/* Pop the obstack selection stack.  */

void
pop_obstacks ()
{
  struct obstack_stack *p = obstack_stack;
  obstack_stack = p->next;

  current_obstack = p->current;
  saveable_obstack = p->saveable;
  expression_obstack = p->expression;
  rtl_obstack = p->rtl;

  obstack_free (&obstack_stack_obstack, p);
}

/* Nonzero if temporary allocation is currently in effect.
   Zero if currently doing permanent allocation.  */

int
allocation_temporary_p ()
{
  return current_obstack != &permanent_obstack;
}

/* Go back to allocating on the permanent obstack
   and free everything in the temporary obstack.

   FUNCTION_END is true only if we have just finished compiling a function.
   In that case, we also free preserved initial values on the momentary
   obstack.  */

void
permanent_allocation (function_end)
     int function_end;
{
  /* Free up previous temporary obstack data */
  obstack_free (&temporary_obstack, temporary_firstobj);
  if (function_end)
    {
      obstack_free (&momentary_obstack, momentary_function_firstobj);
      momentary_firstobj = momentary_function_firstobj;
    }
  else
    obstack_free (&momentary_obstack, momentary_firstobj);
  obstack_free (function_maybepermanent_obstack, maybepermanent_firstobj);
  obstack_free (&temp_decl_obstack, temp_decl_firstobj);

  /* Free up the maybepermanent_obstacks for any of our nested functions
     which were compiled at a lower level.  */
  while (inline_obstacks)
    {
      struct simple_obstack_stack *current = inline_obstacks;
      inline_obstacks = current->next;
      obstack_free (current->obstack, 0);
      free (current->obstack);
      free (current);
    }

  current_obstack = &permanent_obstack;
  expression_obstack = &permanent_obstack;
  rtl_obstack = saveable_obstack = &permanent_obstack;
}

/* Save permanently everything on the maybepermanent_obstack.  */

void
preserve_data ()
{
  maybepermanent_firstobj
    = (char *) obstack_alloc (function_maybepermanent_obstack, 0);
}

void
preserve_initializer ()
{
  struct momentary_level *tem;
  char *old_momentary;

  temporary_firstobj
    = (char *) obstack_alloc (&temporary_obstack, 0);
  maybepermanent_firstobj
    = (char *) obstack_alloc (function_maybepermanent_obstack, 0);

  old_momentary = momentary_firstobj;
  momentary_firstobj
    = (char *) obstack_alloc (&momentary_obstack, 0);
  if (momentary_firstobj != old_momentary)
    for (tem = momentary_stack; tem; tem = tem->prev)
      tem->base = momentary_firstobj;
}

/* Start allocating new rtl in current_obstack.
   Use resume_temporary_allocation
   to go back to allocating rtl in saveable_obstack.  */

void
rtl_in_current_obstack ()
{
  rtl_obstack = current_obstack;
}

/* Start allocating rtl from saveable_obstack.  Intended to be used after
   a call to push_obstacks_nochange.  */

void
rtl_in_saveable_obstack ()
{
  rtl_obstack = saveable_obstack;
}

/* Allocate SIZE bytes in the current obstack
   and return a pointer to them.
   In practice the current obstack is always the temporary one.  */

char *
oballoc (size)
     int size;
{
  return (char *) obstack_alloc (current_obstack, size);
}

/* Free the object PTR in the current obstack
   as well as everything allocated since PTR.
   In practice the current obstack is always the temporary one.  */

void
obfree (ptr)
     char *ptr;
{
  obstack_free (current_obstack, ptr);
}

/* Allocate SIZE bytes in the permanent obstack
   and return a pointer to them.  */

char *
permalloc (size)
     int size;
{
  return (char *) obstack_alloc (&permanent_obstack, size);
}

/* Allocate NELEM items of SIZE bytes in the permanent obstack
   and return a pointer to them.  The storage is cleared before
   returning the value.  */

char *
perm_calloc (nelem, size)
     int nelem;
     long size;
{
  char *rval = (char *) obstack_alloc (&permanent_obstack, nelem * size);
  bzero (rval, nelem * size);
  return rval;
}

/* Allocate SIZE bytes in the saveable obstack
   and return a pointer to them.  */

char *
savealloc (size)
     int size;
{
  return (char *) obstack_alloc (saveable_obstack, size);
}

/* Allocate SIZE bytes in the expression obstack
   and return a pointer to them.  */

char *
expralloc (size)
     int size;
{
  return (char *) obstack_alloc (expression_obstack, size);
}

/* Print out which obstack an object is in.  */

void
print_obstack_name (object, file, prefix)
     char *object;
     FILE *file;
     char *prefix;
{
  struct obstack *obstack = NULL;
  char *obstack_name = NULL;
  struct function *p;

  for (p = outer_function_chain; p; p = p->next)
    {
      if (_obstack_allocated_p (p->function_obstack, object))
	{
	  obstack = p->function_obstack;
	  obstack_name = "containing function obstack";
	}
      if (_obstack_allocated_p (p->function_maybepermanent_obstack, object))
	{
	  obstack = p->function_maybepermanent_obstack;
	  obstack_name = "containing function maybepermanent obstack";
	}
    }

  if (_obstack_allocated_p (&obstack_stack_obstack, object))
    {
      obstack = &obstack_stack_obstack;
      obstack_name = "obstack_stack_obstack";
    }
  else if (_obstack_allocated_p (function_obstack, object))
    {
      obstack = function_obstack;
      obstack_name = "function obstack";
    }
  else if (_obstack_allocated_p (&permanent_obstack, object))
    {
      obstack = &permanent_obstack;
      obstack_name = "permanent_obstack";
    }
  else if (_obstack_allocated_p (&momentary_obstack, object))
    {
      obstack = &momentary_obstack;
      obstack_name = "momentary_obstack";
    }
  else if (_obstack_allocated_p (function_maybepermanent_obstack, object))
    {
      obstack = function_maybepermanent_obstack;
      obstack_name = "function maybepermanent obstack";
    }
  else if (_obstack_allocated_p (&temp_decl_obstack, object))
    {
      obstack = &temp_decl_obstack;
      obstack_name = "temp_decl_obstack";
    }

  /* Check to see if the object is in the free area of the obstack.  */
  if (obstack != NULL)
    {
      if (object >= obstack->next_free
	  && object < obstack->chunk_limit)
	fprintf (file, "%s in free portion of obstack %s",
		 prefix, obstack_name);
      else
	fprintf (file, "%s allocated from %s", prefix, obstack_name);
    }
  else
    fprintf (file, "%s not allocated from any obstack", prefix);
}

void
debug_obstack (object)
     char *object;
{
  print_obstack_name (object, stderr, "object");
  fprintf (stderr, ".\n");
}

/* Return 1 if OBJ is in the permanent obstack.
   This is slow, and should be used only for debugging.
   Use TREE_PERMANENT for other purposes.  */

int
object_permanent_p (obj)
     tree obj;
{
  return _obstack_allocated_p (&permanent_obstack, obj);
}

/* Start a level of momentary allocation.
   In C, each compound statement has its own level
   and that level is freed at the end of each statement.
   All expression nodes are allocated in the momentary allocation level.  */

void
push_momentary ()
{
  struct momentary_level *tem
    = (struct momentary_level *) obstack_alloc (&momentary_obstack,
						sizeof (struct momentary_level));
  tem->prev = momentary_stack;
  tem->base = (char *) obstack_base (&momentary_obstack);
  tem->obstack = expression_obstack;
  momentary_stack = tem;
  expression_obstack = &momentary_obstack;
}

/* Set things up so the next clear_momentary will only clear memory
   past our present position in momentary_obstack.  */

void
preserve_momentary ()
{
  momentary_stack->base = (char *) obstack_base (&momentary_obstack);
}

/* Free all the storage in the current momentary-allocation level.
   In C, this happens at the end of each statement.  */

void
clear_momentary ()
{
  obstack_free (&momentary_obstack, momentary_stack->base);
}

/* Discard a level of momentary allocation.
   In C, this happens at the end of each compound statement.
   Restore the status of expression node allocation
   that was in effect before this level was created.  */

void
pop_momentary ()
{
  struct momentary_level *tem = momentary_stack;
  momentary_stack = tem->prev;
  expression_obstack = tem->obstack;
  /* We can't free TEM from the momentary_obstack, because there might
     be objects above it which have been saved.  We can free back to the
     stack of the level we are popping off though.  */
  obstack_free (&momentary_obstack, tem->base);
}

/* Pop back to the previous level of momentary allocation,
   but don't free any momentary data just yet.  */

void
pop_momentary_nofree ()
{
  struct momentary_level *tem = momentary_stack;
  momentary_stack = tem->prev;
  expression_obstack = tem->obstack;
}

/* Call when starting to parse a declaration:
   make expressions in the declaration last the length of the function.
   Returns an argument that should be passed to resume_momentary later.  */

int
suspend_momentary ()
{
  register int tem = expression_obstack == &momentary_obstack;
  expression_obstack = saveable_obstack;
  return tem;
}

/* Call when finished parsing a declaration:
   restore the treatment of node-allocation that was
   in effect before the suspension.
   YES should be the value previously returned by suspend_momentary.  */

void
resume_momentary (yes)
     int yes;
{
  if (yes)
    expression_obstack = &momentary_obstack;
}

/* Init the tables indexed by tree code.
   Note that languages can add to these tables to define their own codes.  */

void
init_tree_codes ()
{
  tree_code_type = (char **) xmalloc (sizeof (standard_tree_code_type));
  tree_code_length = (int *) xmalloc (sizeof (standard_tree_code_length));
  tree_code_name = (char **) xmalloc (sizeof (standard_tree_code_name));
  bcopy ((char *) standard_tree_code_type, (char *) tree_code_type,
	 sizeof (standard_tree_code_type));
  bcopy ((char *) standard_tree_code_length, (char *) tree_code_length,
	 sizeof (standard_tree_code_length));
  bcopy ((char *) standard_tree_code_name, (char *) tree_code_name,
	 sizeof (standard_tree_code_name));
}

/* Return a newly allocated node of code CODE.
   Initialize the node's unique id and its TREE_PERMANENT flag.
   For decl and type nodes, some other fields are initialized.
   The rest of the node is initialized to zero.

   Achoo!  I got a code in the node.  */