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