[cfe-dev] Default stack alignment for x86 changed

[mats petersson]

To be clear: double does not REQUIRE 8 byte alignment, but on
(reasonably modern, like "Pentium onwards", so ca 1994-5 ish) x86
processors would "prefer" 8-byte alignment for "double" values, since
they can then be read as ONE cycle on a 64-bit bus.

And of course, SSE instructions that aren't specifically designed for
unaligned loads will require a 16-byte alignment. Or does SSE code
automatically modify the alignment criteria for the function?

Further, shouldn't the stack be aligned to "LargestAlignment" or
whatever it is called? Otherwise, any structure alignment will surely
be "lost"?

--
Mats

On 15 January 2015 at 18:38, Smith, Kevin B <kevin.b.smith at intel.com> wrote:
> Although alignof(double) on windows returns 8, the actual minimum stack alignment is still 4.  Here is a source example illustrating
> this.
>
> #include <stdlib.h>
>
> int a = __alignof(double);
>
> extern void crud1(int i, double *p);
>
> void crud(void) {
>   double dummy;
>   crud1(0, &dummy);
> }
>
> Assembly code produced from VS 2012, compiling with cl -Fa -c -O2 crud.c
> _DATA   SEGMENT
> _a      DD      08H
> _DATA   ENDS
> PUBLIC  _crud
> EXTRN   _crud1:PROC
> ; Function compile flags: /Ogtpy
> ;       COMDAT _crud
> _TEXT   SEGMENT
> _dummy$ = -8
> _crud   PROC
> ; File d:\users\kbsmith1\tc_tmp1\crud.c
> ; Line 7
>         sub     esp, 8
> ; Line 9
>         lea     eax, DWORD PTR _dummy$[esp+8]
>         push    eax
>         push    0
>         call    _crud1
> ; Line 10
>         add     esp, 16
>         ret     0
>
> You can see that __alignof(double) produced 8 by the initialization value of a.  You can also see that there is no code at the beginning of function crud to
> align the stack.  So, if it comes in on a 4 byte boundary, it will remain on a 4 byte boundary, and since it subs 8 from esp, if it comes in on an 8 byte boundary
> it will stay on an 8 byte boundary.  Now consider the call to crud1.  This pushes two parameters, and then the call pushes the return address.  So, if the stack
> comes in 8 byte aligned, at the entry to crud1, the stack is now only 4 byte aligned.
>
> For this reason, in windows, although __alignof(double) is 8, it doesn't follow that the value of every double * must be such that the pointer value is 8 byte aligned.
>
> Also, for IA32 on linux, 4 byte minimum stack alignment used to be specified by the Sys V ABI, which is pretty much the only one you can find references to on the web.  However, for quite a number of years, gcc's default on linux is to assure 16 byte stack alignment at function entry, so that every function that used SSE/SSE2 instructions (and might possibly need to spill) didn't have to perform dynamic stack alignment.  In gcc this is controlled by -mpreferred-stack-boundary=num option.  https://gcc.gnu.org/onlinedocs/gcc-3.2/gcc/i386-and-x86-64-Options.html, says the default for this option is 4, implying 16 byte stack alignment.
>
> Kevin Smith
>
> -----Original Message-----
> From: cfe-dev-bounces at cs.uiuc.edu [mailto:cfe-dev-bounces at cs.uiuc.edu] On Behalf Of palparni
> Sent: Thursday, January 15, 2015 9:34 AM
> To: cfe-dev at cs.uiuc.edu
> Subject: Re: [cfe-dev] Default stack alignment for x86 changed
>
> I understand, so the change was made for Unix-based systems in mind.
> Unfortunately the win32 x86 ABI seems to require doubles to be 64-bit
> aligned. Could we perhaps keep the 8-byte alignment only for win32 targets?
>
> Thanks,
> Alpar
>
>
>
> --
> View this message in context: http://clang-developers.42468.n3.nabble.com/Default-stack-alignment-for-x86-changed-tp4043481p4043483.html
> Sent from the Clang Developers mailing list archive at Nabble.com.
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