On Thu, 2018-03-01 at 16:19 -0700, Logan Gunthorpe wrote:
(Switching back to my non-IBM address ...)
On 01/03/18 04:00 PM, Benjamin Herrenschmidt wrote:
> We use only 52 in practice but yes.
> > That's 64PB. If you use need
> > a sparse vmemmap for the entire space it will take 16TB which leaves you
> > with 63.98PB of address space left. (Similar calculations for other
> > numbers of address bits.)
> We only have 52 bits of virtual space for the kernel with the radix
Ok, assuming you only have 52 bits of physical address space: the sparse
vmemmap takes 1TB and you're left with 3.9PB of address space for other
things. So, again, why doesn't that work? Is my math wrong
The big problem is not the vmemmap, it's the linear mapping.