Re: Segfault on the i386 enter instruction

[Chuck Ebbert <76306.1226@compuserve.com>]

In-Reply-To: <44676F42.7080907@aknet.ru>

On Sun, 14 May 2006 21:56:18 +0400, Stas Sergeev wrote:

> Andi Kleen wrote:
> > Handling it like you expect would require to disassemble 
> > the function in the page fault handler and it's probably not 
> > worth doing that for this weird case.
> Just wondering, is this case really that weird?
> In fact, the check against %esp that the kernel
> does, looks strange. I realize that it can catch a
> (very rare) user-space bug of accessing below %esp, but
> other than that it looks redundant (IMHO) and as soon as
> it triggers the false-positives, what is it really good for?

I can't get a SIGSEGV on any native i386 kernel, not even when
running on AMD64.  It only happens on native x86_64 kernels.

Looking at the AMD instruction manual, the pseudo-code for 'enter'
ends with:

RSP.s = RSP - temp_ALLOC_SPACE // Leave "temp_ALLOC_SPACE" free bytes on
                               // the stack
WRITE_MEM.v [SS:RSP.s] = temp_unused // ENTER finishes with a memory write
                                     // check on the final stack pointer,
                                     // but no write actually occurs.
RBP.v = temp_RBP
EXIT


And the Intel manual says:

IF 64-Bit Mode (StackSize = 64)
THEN
        RBP = FrameTemp;
        RSP = RSP - Size;
ELSE IF StackSize = 32
THEN
        EBP = FrameTemp;
        ESP = ESP - Size;
ELSE (* StackSize = 16 *)
        BP = FrameTemp;
        SP = SP - Size;
FI;
END;


Intel says nothing about a write check.  Is that a mistake in the manual
or is that something only AMD64 does, and then only in long mode?


-- 
Chuck

"The x86 isn't all that complex -- it just doesn't make a lot of sense."
                                                        -- Mike Johnson