On 24/10/12 15:34, mind entropy wrote: > On Wed, Oct 24, 2012 at 3:58 PM, David Brown <da...@westcontrol.com> wrote: > >> On 24/10/2012 09:38, mind entropy wrote: >>> On Wed, Oct 24, 2012 at 12:36 PM, Eric Decker <cire...@gmail.com> >>> wrote: >>> >>>> >>>> >>>> On Tue, Oct 23, 2012 at 11:14 PM, mind entropy >>>> <mindentr...@gmail.com>wrote: >>>> >>>>> Hi, >>>>> >>>>> I was reading the mspgcc manual ( >>>>> >>>>> >> http://www.eecs.harvard.edu/~konrad/projects/motetrack/mspgcc-manual-20031127.pdf >>>>> >>>>> >> ) >>>>> and on Pg 39 its written as >>>>> >>>> >>>> I believe there is a more up to date version but I'm not usre where >>>> it lives. Perhaps Peter can point to it? >>>> >>>> >>>>> >>>>> "If you execute while ((long) a & 0x80000l); the program will >>>>> hang, unless ’a’ is declared volatile. So, do it!" >>>>> >>>> With optimization a will be put into a register (actually two >>>> registers since it is a long) and loaded once. >>>> >>>> If a is declared as a volatile, that tells the toolchain that a can >>>> change out from underneath any assumptions made about contents. >>>> This forces the compiler to reload a into a register each >>>> iteration. >>>> >>> >>> Its loaded once during the start of the iteration right? >>> Theoretically if there is no change from outside is there a point in >>> making it volatile? i.e. if that loop block is the only thing in the >>> loop? >>> >> >> If "a" is not volatile, the compiler will generate code that is like this: >> >> long localCopyOfA = a; >> if (localCopyOfA & 0x800001) { >> while (true) ; >> } >> >> If that is what you meant to write, then it's fine to have "a" >> non-volatile. But I /really/ doubt that this is what you want. >> >> The loop is completely useless unless "a" is modified from outside in >> some way - from an interrupt function (the typical case), via hardware >> (maybe "a" is a register, or perhaps modified by DMA), from another RTOS >> thread, or even via a debugger (such constructs can sometimes be useful >> in testing and debugging). In all these cases, you have to let the >> compiler know that "a" may be modified externally - and you do that by >> declaring it "volatile". >> >> > Yes I realized it later. The code I have below would not need to have > volatile right since > the change is not made externally from an interrupt etc (Assuming that code > is the only > code in the program) > > unsigned long a = 0L; > while(a < 10L) { > a++; > } > >
I've had a look at the pdf you linked. While a lot of the information there is useful, there is also a lot that is out of date - be careful about relying too much on the details here. The "Tips and Tricks" chapter you mentioned contains some useful advice, but also a number of mangled or plain incorrect points. Much of them could be summarised as "learn to write correct C". ------------------------------------------------------------------------------ Everyone hates slow websites. So do we. Make your web apps faster with AppDynamics Download AppDynamics Lite for free today: http://p.sf.net/sfu/appdyn_sfd2d_oct _______________________________________________ Mspgcc-users mailing list Mspgcc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/mspgcc-users
