Re: [PATCH] Linux Kernel Markers 0.13 for 2.6.17

From: Mathieu Desnoyers
Date: Tue Sep 26 2006 - 14:04:38 EST


Hi,

Ok, so as far as I can see, we can only control the execution flow by modifying
values in the output list of the asm.

Do you think the following would work ?


#define MARK_JUMP(name, format, args...) \
do { \
char condition; \
asm volatile( ".section .markers, \"a\";\n\t" \
".long 0f;\n\t" \
".previous;\n\t" \
"0:\n\t" \
"movb $0,%1;\n\t" \
: "+m" (__marker_sequencer), \
"=r" (condition) : ); \
if(unlikely(condition)) { \
MARK_CALL(name, format, ## args); \
} \
} while(0)

The jump is left to gcc, we only modify an immediate value (a byte) to change
the selection. The is no memory load involved on the fast path :

...
6: b0 00 mov $0x0,%al
8: 84 c0 test %al,%al
a: 75 07 jne 13 <my_open+0x13>
c: b8 ff ff ff ff mov $0xffffffff,%eax
11: c9 leave
12: c3 ret
13: b8 01 00 00 00 mov $0x1,%eax
18: e8 fc ff ff ff call 19 <my_open+0x19>
1d: c7 44 24 0c 00 00 00 movl $0x0,0xc(%esp)
24: 00
25: c7 44 24 08 06 00 00 movl $0x6,0x8(%esp)
2c: 00
2d: c7 44 24 04 02 00 00 movl $0x2,0x4(%esp)
34: 00
35: c7 04 24 0c 00 00 00 movl $0xc,(%esp)
3c: ff 15 94 00 00 00 call *0x94
42: b8 01 00 00 00 mov $0x1,%eax
47: e8 fc ff ff ff call 48 <my_open+0x48>
4c: eb be jmp c <my_open+0xc>


Mathieu

* Jeremy Fitzhardinge (jeremy@xxxxxxxx) wrote:
> Mathieu Desnoyers wrote:
> >>Mathieu Desnoyers wrote:
> >>
> >>>To protect code from being preempted, the macros preempt_disable and
> >>>preempt_enable must normally be used. Logically, this macro must make
> >>>sure gcc
> >>>doesn't interleave preemptible code and non-preemptible code.
> >>>
> >>>
> >>No, it only needs to prevent globally visible side-effects from being
> >>moved into/out of preemptable blocks. In practice that means memory
> >>updates (including the implicit ones that calls to external functions
> >>are assumed to make).
> >>
> >>
> >>>Which makes me think that if I put barriers around my asm, call, asm
> >>>trio, no
> >>>other code will be interleaved. Is it right ?
> >>>
> >>>
> >>No global side effects, but code with local side effects could be moved
> >>around without changing the meaning of preempt.
> >>
> >>For example:
> >>
> >> int foo;
> >> extern int global;
> >>
> >> foo = some_function();
> >>
> >> foo += 42;
> >>
> >> preempt_disable();
> >> // stuff
> >> preempt_enable();
> >>
> >> global = foo;
> >> foo += other_thing();
> >>
> >>Assume here that some_function and other_function are extern, and so gcc
> >>has no insight into their behaviour and therefore conservatively assumes
> >>they have global side-effects.
> >>
> >>The memory barriers in preempt_disable/enable will prevent gcc from
> >>moving any of the function calls into the non-preemptable region. But
> >>because "foo" is local and isn't visible to any other code, there's no
> >>reason why the "foo += 42" couldn't move into the preempt region.
> >>
> >
> >I am not sure about this last statement. The same reference :
> >http://developer.apple.com/documentation/DeveloperTools/gcc-4.0.1/gcc/Extended-Asm.html
> >
> (This is pretty old, and this is an area which changes quite a lot. You
> should refer to something more recent;
> http://www.cims.nyu.edu/cgi-systems/info2html?/usr/local/info(gcc)Top
> for example, though in this case the quoted text looks the same.)
>
> >I am just wondering how gcc can assume that I will not modify variables on
> >the
> >stack from within a function with a memory clobber. If I would like to do
> >some
> >nasty things in my assembly code, like accessing directly to a local
> >variable by
> >using an offset from the stack pointer, I would expect gcc not to relocate
> >this
> >local variable around my asm volatile memory clobbered statement, as it
> >falls
> >under the category "access memory in an unpredictable fashion".
> >
>
> That not really what it means. gcc is free to put local variables in
> memory or register, and unless you pass the local to your asm as a
> parameter, your code has no way of knowing how to find the current
> location of the local. You could trash your stack frame from within the
> asm if you like, but I don't think gcc is under any obligation to behave
> in a deterministic way if you do.
>
> "Unpredictable" in this case means that the memory modified isn't easily
> specified as a normal asm parameter. For example, if you have an asm
> which does a memset(), the modified memory's size is a runtime variable
> rather than a compile-time constant. Or perhaps your asm follows a
> linked list and modifies memory as it traverses the list.
>
>
> J
>
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