On Thu, 16 Sep 2010 19:51:33 -0700 Tabi Timur-B04825 <b04...@freescale.com> wrote:
> Jaggi Manish-B10520 wrote: > > Most of the read/write functions do this way > > 0:+ int sw = set_mux_to_lbc(); > > 1:+ > > 2:+ ret = __raw_readl(addr); > > 3:+ if (sw) > > 4:+ set_mux_to_diu(); > > > > compiler might reorder 2 before 3 and 0, read / write wont have any > > issue , right ? > > Really? I don't see why the compiler would do that. That would be a bug, > IMHO. Only if set_mux_to_lbc() is non-inline, and the compiler has no visibility into what it does. But even then, the hardware is free to reorder. That's why we have non-raw I/O accessors. > Can you explain why you think it will make such a reordering? My > understanding is that the compiler reorders things that don't have any side > effects. The compiler can reorder anything where the reordering won't affect the results of single-threaded execution with no interrupts, well-behaved memory (no device registers, bank switches, etc), no violations of aliasing rules, etc. -Scott _______________________________________________ U-Boot mailing list U-Boot@lists.denx.de http://lists.denx.de/mailman/listinfo/u-boot
