[U-Boot-Users] [Q] memtest doubt
Jerry Van Baren
gerald.vanbaren at ge.com
Thu Feb 7 14:34:16 CET 2008
Guennadi Liakhovetski wrote:
> Hi all,
>
> looking through the common/cmd_mem.c::do_mem_mtest() function, I couldn't
> understand the following place:
>
> addr_mask = ((ulong)end - (ulong)start)/sizeof(vu_long);
> ...
> for (offset = 1; (offset & addr_mask) != 0; offset <<= 1) {
> start[offset] = pattern;
> }
>
> why (offset & addr_mask) != 0 and not just offset < addr_mask? Suppose
>
> end = 0xbf;
> start = 0;
>
> addr_mask = 0x2f;
>
> The loop will iterate over offset = 1, 2, 4, 8, and on 0x10 it will abort
> and 0x10 and 0x20 will stay untested. Whereas if we just had "offset <
> addr_mask" it would just function correctly, wouldn't it? Yes, I do
> realise, that it is at least unusual to set the end address to anything
> other than start address + ((1 << x) - 1), but still.
>
> Thanks
> Guennadi
Hi Guennadi,
The address test is stepping through the address lines
0x01, 0x02, 0x04, 0x08, 0x10, 0x20
Your end of 0xBF with a mask of 0x2F indicates that the address lines
0x10, 0x40, and 0x80 are not present (even though address line 0x80
looks like it is part of the test since 0xBF includes 0x80 - but it
isn't tested).
This is nonsensical with respect to what the address line test is
testing (address lines!) and how it is testing them - by stepping
through the address lines and looking for inadvertent overlapped memory
accesses. Address lines are inherently powers of two, skipping certain
ones of them doesn't make much sense, and ending not on a power of two
(minus one) doesn't make any sense at all.
I fail to see what your change would benefit as an address line test.
If there is a benefit, you will have to rewrite the address line test
because having a contiguous mask and a ((2^n) - 1) end is all
fundamental to how the address line test works, is fundamental to what
address lines are, and is based on the symptoms that are observable when
an address line fails. Simply using an arbitrary end address and a
funky mask will cause the current test to fail and likely will cause
some real failures to be undetected.
Best regards,
gvb
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