James Georgas wrote:
> That's the vmalloc address space. You only get 896 MB in the NORMAL
> zone on i386, to leave room for vmalloc. If you don't like it, go
64-bit.
>
> -- Chris
I like it fine. I just didn't understand it. Thanks for answering.
So, basically, the vmalloc
James Georgas wrote:
That's the vmalloc address space. You only get 896 MB in the NORMAL
zone on i386, to leave room for vmalloc. If you don't like it, go
64-bit.
-- Chris
I like it fine. I just didn't understand it. Thanks for answering.
So, basically, the vmalloc address
James C. Georgas wrote:
I'm not sure I understand how the kernel calculates the amount of
physical RAM it can map during the boot process.
I've quoted two blocks of kernel messages below, one for a kernel with
NOHIGHMEM and another for a kernel with HIGHMEM4G.
If I do the math on the BIOS
I'm not sure I understand how the kernel calculates the amount of
physical RAM it can map during the boot process.
I've quoted two blocks of kernel messages below, one for a kernel with
NOHIGHMEM and another for a kernel with HIGHMEM4G.
If I do the math on the BIOS provided physical RAM map,
I'm not sure I understand how the kernel calculates the amount of
physical RAM it can map during the boot process.
I've quoted two blocks of kernel messages below, one for a kernel with
NOHIGHMEM and another for a kernel with HIGHMEM4G.
If I do the math on the BIOS provided physical RAM map,
James C. Georgas wrote:
I'm not sure I understand how the kernel calculates the amount of
physical RAM it can map during the boot process.
I've quoted two blocks of kernel messages below, one for a kernel with
NOHIGHMEM and another for a kernel with HIGHMEM4G.
If I do the math on the BIOS
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