On 2/25/07, Jarod Wang <mailtojarod@xxxxxxxxx> wrote:
2007/2/25, Mansha Linux <mansha.linux@xxxxxxxxx>:
> hi all,
>
> For a 32-bit machine, there is total 4GB of memory(Physiacl addresses). So
> in that 1 GB for Kernel and 3GB for User Space.
The kernel (on the x86 architecture, in the default configuration)
splits the 4-GB *virtual address space* between user-space and the
kernel; the same set of mappings is used in both contexts. A typical
split dedicates 3 GB to user space, and 1 GB for kernel space.
> So, kernel won't access that 3GB of memory, right?
> All the kernel code is exist in 1 GB of memory.... CMIIW
All the kernel code is linked using 0xC0000000(3GB, also virtual
address) as the base address.
> So, my doubt is the
> DMA(0-16MB),NORMAL(16-896MB),HIGH(896MB-) memory divided
> within this 4GB of memory ,right?
Within 4GB physical addresses space.
> if we allocate the pages using alloc_pages, from which region the pages are
> allocated for the kernel modules. is it from Physical RAM memory or from 1GB
> of kernel space ??
From physical memory.
--
Best regards,
Jarod Wang
----
Wuhan University of Technology, Wuhan, P. R. China
Jarod thanks for ur reply
>The kernel (on the x86 architecture, in the default configuration)
>splits the 4-GB *virtual address space* between user-space and the
>kernel; the same set of mappings is used in both contexts. A typical
>split dedicates 3 GB to user space, and 1 GB for kernel space.
Q->it means for a 32-bit machine all 4GB space is Virtual Addresses not Physical addresses ?
--
regards,
Mansha
--
regards,
Mansha
