[...]
I think this structure is simple enough to be fast.
For memory usage:
* Each entry of the first array describes 1GB of memory. An entry is 32
bits. 16KB for the first array can describe 2**12 * 2**30 = 2**42 B of
memory. (Dan's machine physical memory is bellow 2**40).
* I think 1GB of granule is good enough, unless you have a machine with
very small DIMM. In this case, we can use 512MB or 256MB instead of
1GB. * 1GB is 2**16 to 2**18 pages. Thus, the offset may be 18 bits
and the length 14 bits (to be multiplied by 4).
As a conclusion, the memory footprint is *very* small, maybe too small ?
memmap related macros must be rewritten.
Tristan.
Hugh, I forgot we need also reverse mapping. I now understand the root of
virtual mem map!
Tristan.