Drop excess size used for run time allocated stack variables.

* get_dynamic_stack_size is passed a SIZE of a data block (which is
  allocated elsewhere), the SIZE_ALIGN of the SIZE (i.e. the alignment
  of the underlying memory units (e.g. 32 bytes split into 4 times 8
  bytes = 64 bit alignment) and the REQUIRED_ALIGN of the data portion
  of the allocated memory.

* Assuming the function is called with SIZE = 2, SIZE_ALIGN = 8 and
  REQUIRED_ALIGN = 64 it first adds 7 bytes to SIZE -> 9.  This is
  what is needed to have two bytes 8-byte-aligned at some memory
  location without any known alignment.

* Finally round_push is called to round up SIZE to a multiple of the
  stack slot size.

The key to understanding this is that the function assumes that
STACK_DYNMAIC_OFFSET is completely unknown at the time its called
and therefore it does not make assumptions about the alignment of
STACKPOINTER + STACK_DYNMAIC_OFFSET.  The latest patch simply
hard-codes that SP + SDO is supposed to be aligned to at least
stack slot size (and does that in a very complicated way).  Since
there is no guarantee that this is the case on all targets, the
patch is broken.  It may miscalculate a SIZE that is too small in
some cases.

However, on many targets there is some guarantee about the
alignment of SP + SDO even if the actual value of SDO is unknown.
On s390x it's always 8-byte-aligned (stack slot size).  So the
right fix should be to add knowledge about the target's guaranteed
alignment of SP + SDO to the function.  I'm right now testing a
much simpler patch that uses
REGNO_POINTER_ALIGN(VIRTUAL_STACK_DYNAMIC_REGNUM) as the
alignment.


gcc/ChangeLog:

2016-08-23  Dominik Vogt  <vogt@linux.vnet.ibm.com>

	* explow.c (get_dynamic_stack_size): Take known alignment of stack
	pointer + STACK_DYNAMIC_OFFSET into account when calculating the
	size needed.  Correct a typo in a comment.

From-SVN: r239688

0 files changed, 0 insertions, 0 deletions