ia64: fix small struct return

This change fixes libffi.call/struct10.c failure on ia64:
FAIL: libffi.call/struct10.c -W -Wall -Wno-psabi -O0 execution test

.Lst_small_struct handles returns for structs less than 32 bytes
(following ia64 return value ABI [1]). Subroutine does roughly the
following:

```
    mov [sp+0]  = r8
    mov [sp+8]  = r9
    mov [sp+16] = r10
    mov [sp+24] = r11
    memcpy(destination, source=sp, 12);
```

The problem: ia64 ABI guarantees that top 16 bytes of stack are
scratch space for callee function. Thus it can clobber it. [1]
says (7.1 Procedure Frames):
"""
* Scratch area. This 16-byte region is provided as scratch storage
  for procedures that are called by the current procedure. Leaf
  procedures do not need to allocate this region. A procedure may
  use the 16 bytes at the top of its own frame as scratch memory,
  but the contents of this area are not preserved by a procedure call.
"""

In our case 16 top bytes are clobbered by a PLT resolver when memcpy()
is called for the first time. As a result memcpy implementation reads
already clobbered data frop top of stack.

The fix is simple: allocate 16 bytes of scrats space prior to memcpy()
call.

[1]: https://www.intel.com/content/dam/www/public/us/en/documents/guides/itanium-software-runtime-architecture-guide.pdf

Bug: https://bugs.gentoo.org/634190
Signed-off-by: Sergei Trofimovich <slyfox@gentoo.org>

Author: Sergei Trofimovich

src/ia64/unix.S

@@ -175,7 +175,6 @@ ffi_call_unix:
 	;;
 
 .Lst_small_struct:
-	add	sp = -16, sp
 	cmp.lt	p6, p0 = 8, in3
 	cmp.lt	p7, p0 = 16, in3
 	cmp.lt	p8, p0 = 24, in3
@@ -191,6 +190,12 @@ ffi_call_unix:
 (p8)	st8	[r18] = r11
 	mov	out1 = sp
 	mov	out2 = in3
+	;;
+	// ia64 software calling convention requires
+	// top 16 bytes of stack to be scratch space
+	// PLT resolver uses that scratch space at
+	// 'memcpy' symbol reolution time
+	add	sp = -16, sp
 	br.call.sptk.many b0 = memcpy#
 	;;
 	mov	ar.pfs = loc0