Question on virtual memory layout: lowmem with memory hole

[linux@arm.linux.org.uk (Russell King - ARM Linux)]

On Thu, Aug 25, 2011 at 09:35:07AM +0530, Pedanekar, Hemant wrote:
> E.g., on OMAP3 with mem=32M at 0x80000000 mem=8M at 0x87800000
> 
> (CASE 1)
>  Memory: 32MB 8MB = 40MB total
>  Memory: 28408k/28408k available, 12552k reserved, 0K highmem
>  Virtual kernel memory layout:
>      vector  : 0xffff0000 - 0xffff1000   (   4 kB)
>      fixmap  : 0xfff00000 - 0xfffe0000   ( 896 kB)
>      DMA     : 0xffc00000 - 0xffe00000   (   2 MB)
>      vmalloc : 0xc8800000 - 0xf8000000   ( 760 MB)
>      lowmem  : 0xc0000000 - 0xc8000000   ( 128 MB)
>      modules : 0xbf000000 - 0xc0000000   (  16 MB)
>        .text : 0xc0008000 - 0xc05ac2c8   (5777 kB)
>        .init : 0xc05ad000 - 0xc05f8780   ( 302 kB)
>        .data : 0xc05fa000 - 0xc06838d0   ( 551 kB)
>         .bss : 0xc06838f4 - 0xc0bd8a14   (5461 kB)
> 
> Is this expected? 88MB space between two 'mem's seems to have lost.

Yes.  The memory layout lines gives an overview of the virtual memory
address space _regions_.

What it's saying is that the virtual addresses from 0xc0000000 - 0xc8000000
are used for lowmem.  That may not be fully populated, but that's what
the address range is reserved for.

> This also
> means vmalloc space is lower compared to when a single mem=40M is passed.

Huh.  Either your maths is wrong or...

Here's case 1:
>      vmalloc : 0xc8800000 - 0xf8000000   ( 760 MB)
And case 2:
>      vmalloc : 0xc3000000 - 0xf8000000   ( 848 MB)

Looks to me like case 1, vmalloc space is _higher_ not _lower_.  That's
expected because you told the kernel it had more memory in case 1.