Question on virtual memory layout: lowmem with memory hole

Hello,

While checking with memory hole configuration on OMAP3/TI816X (passing mem=x at y),
I see that the whole range from memory start till last bank including the hole
is labelled as 'lowmem' when kernel prints "Virtual kernel memory layout" during
boot.

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

E.g., with mem=40M

(CASE 2)
 Memory: 40MB = 40MB total
 Memory: 28412k/28412k available, 12548k reserved, 0K highmem
 Virtual kernel memory layout:
     vector  : 0xffff0000 - 0xffff1000   (   4 kB)
     fixmap  : 0xfff00000 - 0xfffe0000   ( 896 kB)
     DMA     : 0xffc00000 - 0xffe00000   (   2 MB)
     vmalloc : 0xc3000000 - 0xf8000000   ( 848 MB)
     lowmem  : 0xc0000000 - 0xc2800000   (  40 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)

Any way we can get virtual layout same in CASE 1 as in CASE 2?

Thanks.

   Hemant

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.

Russell King - ARM Linux wrote on Thursday, August 25, 2011 3:53 PM:

On Thu, Aug 25, 2011 at 09:35:07AM +0530, Pedanekar, Hemant wrote:

quoted

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.

So, larger the hole, more address space will be unusable - is that correct?
 

quoted

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:

quoted

     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.

Sorry, my mistake - I actually meant "vmalloc space is _smaller_ compared to
when a single mem=40M is passed" though the actual physical RAM available is
same in both the cases.

Thanks.

   Hemant

On Fri, Aug 26, 2011 at 12:06:20AM +0530, Pedanekar, Hemant wrote:

Russell King - ARM Linux wrote on Thursday, August 25, 2011 3:53 PM:

quoted

On Thu, Aug 25, 2011 at 09:35:07AM +0530, Pedanekar, Hemant wrote:

quoted

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.

So, larger the hole, more address space will be unusable - is that correct?

Correct.

quoted

quoted

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:

quoted

     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.

Sorry, my mistake - I actually meant "vmalloc space is _smaller_ compared to
when a single mem=40M is passed" though the actual physical RAM available is
same in both the cases.

Again, that's expected.

We require a 1:1 relationship between virtual and physical addresses for
efficiency - having non-linear translation means we'd need a lookup table,
and as these translations are heavily used, that will impact performance.

Russell King - ARM Linux wrote on Friday, August 26, 2011 12:22 AM:

On Fri, Aug 26, 2011 at 12:06:20AM +0530, Pedanekar, Hemant wrote:

quoted

Russell King - ARM Linux wrote on Thursday, August 25, 2011 3:53 PM:

quoted

On Thu, Aug 25, 2011 at 09:35:07AM +0530, Pedanekar, Hemant wrote:

quoted

E.g., on OMAP3 with mem=32M at 0x80000000 mem=8M at 0x87800000

[...]

quoted

So, larger the hole, more address space will be unusable - is that correct?

Correct.

quoted

quoted

quoted

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:

quoted

     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.

Sorry, my mistake - I actually meant "vmalloc space is _smaller_ compared
to when a single mem=40M is passed" though the actual physical RAM
available is same in both the cases.

Again, that's expected.

We require a 1:1 relationship between virtual and physical addresses for
efficiency - having non-linear translation means we'd need a lookup table,
and as these translations are heavily used, that will impact
performance.

Thanks for all the clarifications.

   Hemant