On 2024-04-16 Mike Rapoport wrote:
> On Tue, Apr 16, 2024 at 06:00:29PM +0100, Matthew Wilcox wrote:
> > On Tue, Apr 16, 2024 at 07:31:54PM +0300, Mike Rapoport wrote:
> > > > - if (!IS_ENABLED(CONFIG_64BIT)) {
> > > > - max_mapped_addr = __pa(~(ulong)0);
> > > > - if (max_mapped_addr == (phys_ram_end - 1))
> > > > - memblock_set_current_limit(max_mapped_addr - 4096);
> > > > - }
> > > > + memblock_reserve(__pa(-PAGE_SIZE), PAGE_SIZE);
> > >
> > > Ack.
> >
> > Can this go to generic code instead of letting architecture maintainers
> > fall over it?
>
> Yes, it's just have to happen before setup_arch() where most architectures
> enable memblock allocations.
This also works, the reported problem disappears.
However, I am confused about one thing: doesn't this make one page of
physical memory inaccessible?
Is it better to solve this by setting max_low_pfn instead? Then at
least the page is still accessible as high memory.
Best regards,
Nam
diff --git a/arch/riscv/mm/init.c b/arch/riscv/mm/init.c
index fa34cf55037b..6e3130cae675 100644
--- a/arch/riscv/mm/init.c
+++ b/arch/riscv/mm/init.c
@@ -197,7 +197,6 @@ early_param("mem", early_mem);
static void __init setup_bootmem(void)
{
phys_addr_t vmlinux_end = __pa_symbol(&_end);
- phys_addr_t max_mapped_addr;
phys_addr_t phys_ram_end, vmlinux_start;
if (IS_ENABLED(CONFIG_XIP_KERNEL))
@@ -235,23 +234,9 @@ static void __init setup_bootmem(void)
if (IS_ENABLED(CONFIG_64BIT))
kernel_map.va_pa_offset = PAGE_OFFSET - phys_ram_base;
- /*
- * memblock allocator is not aware of the fact that last 4K bytes of
- * the addressable memory can not be mapped because of IS_ERR_VALUE
- * macro. Make sure that last 4k bytes are not usable by memblock
- * if end of dram is equal to maximum addressable memory. For 64-bit
- * kernel, this problem can't happen here as the end of the virtual
- * address space is occupied by the kernel mapping then this check must
- * be done as soon as the kernel mapping base address is determined.
- */
- if (!IS_ENABLED(CONFIG_64BIT)) {
- max_mapped_addr = __pa(~(ulong)0);
- if (max_mapped_addr == (phys_ram_end - 1))
- memblock_set_current_limit(max_mapped_addr - 4096);
- }
-
min_low_pfn = PFN_UP(phys_ram_base);
- max_low_pfn = max_pfn = PFN_DOWN(phys_ram_end);
+ max_pfn = PFN_DOWN(phys_ram_end);
+ max_low_pfn = min(max_pfn, PFN_DOWN(__pa(-PAGE_SIZE)));
high_memory = (void *)(__va(PFN_PHYS(max_low_pfn)));
dma32_phys_limit = min(4UL * SZ_1G, (unsigned long)PFN_PHYS(max_low_pfn));
On Wed, Apr 17, 2024 at 12:36:39AM +0200, Nam Cao wrote:
>
> However, I am confused about one thing: doesn't this make one page of
> physical memory inaccessible?
So are these riscv32 systems really having multiple terabytes of
memory? Why is this page in the physical memory map in the first
place?
Inquiring minds what to know,
- Ted