one question about pgd allocation

one question about pgd allocation

[liyu@WAN]

hi LKML:

    I am reading kernel code about memory management.

    on i386 platform, the function pgd_alloc() alloc one page table 
directory,
every directory entry is one PMD address. that is OK.

pgd_t *pgd_alloc(struct mm_struct *mm)
{
    int i;
    pgd_t *pgd = kmem_cache_alloc(pgd_cache, GFP_KERNEL);

    if (PTRS_PER_PMD == 1 || !pgd)
        return pgd;

    for (i = 0; i < USER_PTRS_PER_PGD; ++i) {
        pmd_t *pmd = kmem_cache_alloc(pmd_cache, GFP_KERNEL);
        if (!pmd)
            goto out_oom;
        set_pgd(&pgd[i], __pgd(1 + __pa(pmd)));
    }
    return pgd;

out_oom:
    for (i--; i >= 0; i--)
        kmem_cache_free(pmd_cache, (void *)__va(pgd_val(pgd[i])-1));
    kmem_cache_free(pgd_cache, pgd);
    return NULL;
}

    My question is the statement:

        set_pgd(&pgd[i], __pgd(1 + __pa(pmd)));

    why here is not :

        set_pgd(&pgd[i], __pgd(__pa(pmd)));

    I can not understand this. I search intel developer mannal. but 
nothing to help.
Who can tell me why?


    thank in advanced.

                                                          liyu
                                                                2005/7/1

Re: one question about pgd allocation

[Andi Kleen]

"liyu@WAN" <liyu@ccoss.com.cn> writes:
> 
>     I can not understand this. I search intel developer mannal. but
> nothing to help.
> Who can tell me why?

bit 0 is the present bit which needs to be set on the PGD, otherwise
the CPU will ignore it. A clearer way to write this would have been

        __pgd(__pa(pmd) | _PAGE_PRESENT)

-Andi