On Thu, Nov 04, 2021 at 01:15:46PM +0100, Emil Renner Berthing wrote: > On Tue, 2 Nov 2021 at 22:17, Emil Renner Berthing <kernel@xxxxxxxx> wrote: ... > I'd really like to understand your reasoning here. As far as I can > tell reading 2 adjacent 32bit registers with a 64bit read as you're > proposing is exactly what would cause endian issues. Eg. on little > endian you'd get reg0 | reg1 << 32 whereas on big-endian you'd get > reg0 << 32 | reg1. Nope, it won't. The endianess is a property of both CPU and device. The I/O accessors, such as readl()/writel() and iowrtieXX()/ioreadXX() are _always_ LE. So, writeq() will properly put bits to their places in case device is LE. And most devices are LE (or should be). Of course there are cases, but then you have to specify them explicitly. My motive here is simple as that the device is definitely a set of a few 128-bit bitmaps (in registers) and using bitmap _is_ representing hardware in the kernel. Using something else will deviate from that (maybe not too far, but still...). -- With Best Regards, Andy Shevchenko
