Messages in this thread |  | | | Date | Mon, 25 Sep 2006 22:03:28 -0700 | | From | Jeremy Fitzhardinge <> | | Subject | Re: [PATCH] Linux Kernel Markers 0.13 for 2.6.17 |
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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.
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