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x86,boot: Don't reserve regular memory region

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It's incorrect to add a region in boot_state.mem_p_regs to the reserved
region list. The reserved region list's purpose is for restricting the
creation of any untyped objects. This error was being masked by a second
call to reserve_region with a partially overlapping region that caused
the memory region to get actually turned into regular untypeds, but the
other reserved region to still get turned into device untypeds.
Now the range [0x0, 0x100000] isn't being added to a reserved region and
will still be turned into device untypeds.
This change shouldn't change the untypeds given to userlevel.

Signed-off-by: Kent McLeod <kent@kry10.com>
This commit is contained in:
Kent McLeod 2021-09-21 18:43:03 +10:00 committed by Kent McLeod
parent 64a14f8fe7
commit bf7c1aa37b
2 changed files with 5 additions and 2 deletions
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5
src/arch/x86/kernel/boot.c
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@ -72,7 +72,10 @@ BOOT_CODE static bool_t arch_init_freemem(p_region_t ui_p_reg,
mem_p_regs_t *mem_p_regs,
word_t extra_bi_size_bits)
{
ui_p_reg.start = 0;
// Extend the reserved region down to include the base of the kernel image.
// KERNEL_ELF_PADDR_BASE is the lowest physical load address used
// in the x86 linker script.
ui_p_reg.start = KERNEL_ELF_PADDR_BASE;
reserved[0] = paddr_to_pptr_reg(ui_p_reg);
return init_freemem(mem_p_regs->count, mem_p_regs->list, MAX_RESERVED,
reserved, it_v_reg, extra_bi_size_bits);

2
src/arch/x86/kernel/boot_sys.c
View file

@ -213,7 +213,7 @@ static BOOT_CODE bool_t add_mem_p_regs(p_region_t reg)
printf("Adding physical memory region 0x%lx-0x%lx\n", reg.start, reg.end);
boot_state.mem_p_regs.list[boot_state.mem_p_regs.count] = reg;
boot_state.mem_p_regs.count++;
return reserve_region(reg);
return true;
}
/*