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?