Dear Heinrich, In message <[email protected]> you wrote: > > > Are you sure? How is this supposed to work? > > > > Running it with start = 0 and size = 0xC0000000 it will test the > > memory locations > > > > 0x80000000 > > 0xA0000000 > > 0xB0000000 > > 0xB8000000 > > 0xBC000000 > > 0xBE000000 > > 0xBF000000 > > 0xBF800000 > > 0xBFC00000 > > 0xBFE00000 > > 0xBFF00000 > > 0xBFF80000 > > 0xBFFC0000 > > 0xBFFE0000 > > 0xBFFF0000 > > 0xBFFF8000 > > 0xBFFFC000 > > 0xBFFFE000 > > 0xBFFFF000 > > 0xBFFFF800 > > 0xBFFFFC00 > > 0xBFFFFE00 > > 0xBFFFFF00 > > 0xBFFFFF80 > > 0xBFFFFFC0 > > 0xBFFFFFE0 > > 0xBFFFFFF0 > > 0xBFFFFFF8 > > 0xBFFFFFFC > > 0xBFFFFFFE > > 0xBFFFFFFF > > The last accessible byte is at 0xBFFFFFFF which matches (start + size - > 1) in your example. > > The difference to the current logic is that it does not require start or > size to be power of two.
Yes, I know, but your test strategy makes no sense. You have to check memory at the power-of-two boudanries to find out how big the ram actually is and whether your memory accesses may be mirrored at other locations in the address room. You are just testing the end of the expected size in some way for which I cannot find a good explanation. > If the size input is larger than the actually accessible memory size, > you will get the actual memory size, e.g. lets assume that the last > accessible address is 0x3333: Please understand thtat this function is not a plain stupid size test, but there is more intelligece in it. We have seen many cases where for example the memory controller was set up for 256 MiB address room, but only 64 MiB of memory were installed. The current version of the test reliably detects that there are only 64 MiB of memory presnet, clearly indicating the problem. I think your test would happyly report to see 256 MiB of memory and miss the fact that this is only one of the 4 mirrors of the real memory - and of course any attempt to run Linux on such a system would result in funny crashes... > Now the algorithm returns that the last accessible memory location is > 0x3332. Come on. The purpose of this function as it was designed is to find the size of installed memory in systems where this may change because there may be several hardware configurations. And memory chips tend to come in sizes that are a power of two. And we not only test the size, we also verify basic functionality (including errors like shorts of breaks in the data and address lanes), and this really fast so it can be kept in the production version. You drop all this very practical functionality for a feature which is of hypothetical use at best? > The algorithm runs in O(log(UINT_MAX)) which is the same time complexity > as for the current algorithm. Yes, but it makes absolutely no sense to me. Best regards, Wolfgang Denk -- DENX Software Engineering GmbH, Managing Director: Wolfgang Denk HRB 165235 Munich, Office: Kirchenstr.5, D-82194 Groebenzell, Germany Phone: (+49)-8142-66989-10 Fax: (+49)-8142-66989-80 Email: [email protected] I can't say I've ever been lost, but I was bewildered once for three days. - Daniel Boone (Attributed)
