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I have a large matrix of values that takes up about 2GB of RAM.

I need to form a copy of this matrix, then the original can be swapped out to disk, to be loaded later. The contents of this matrix are important. Computing it initially is expensive, so you cannot easily throw it away and re-create it. It is faster to drop the matrix to disk, and then re-load it from disk, than it is to re-compute it from scratch.

Is there an easier or better way to designate a section of memory to be temporarily put on disk until next access than what I have, which is:

when the resource (2GB matrix) is not needed

  • open a file
  • write the file to disk
  • free the memory

when the resource is needed

  • open file
  • read in matrix
  • delete file from disk

I came across File mapping But I'm not sure this is the right thing to use

bobobobo
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    Does the matrix need to survive program restarts? – Branko Dimitrijevic Jun 28 '12 at 21:44
  • Clearly you are using a 64-bit operating system. It is already very good at figuring out when parts of your virtual memory mapping are cold and can be swapped out with little danger of slowing down your program. Most any kind of operation you do on a matrix touches *all* of the pages, multiplication being the common one. Trying to help will only slow it down. If the matrix is sparse then you do have options, not indicated in the question. – Hans Passant Jun 28 '12 at 22:12
  • @HansPassant: My understanding is that he's performing operations on the copy, probably keeping the original to initialize a new copy when it's time to perform the next set of calculations. The original will likely only be touched while performing the copy, while the copy will likely be frequently and extensively accessed. – Eric J. Jun 28 '12 at 22:15
  • No, the matrix doesn't need to survive program restarts – bobobobo Jul 01 '12 at 22:56

2 Answers2

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Have a look at Memory Mapped Files.

Memory-mapped files (MMFs) offer a unique memory management feature that allows applications to access files on disk in the same way they access dynamic memory—through pointers.

The operating system will very efficiently swap portions of the original matrix to/from disk.

Eric J.
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Assuming the matrix doesn't need to survive program restarts, compile your application as 64-bit and just leave the matrix in memory. The OS will automatically swap-out the least-used memory pages when under memory pressure.

However, even on a mildly modern hardware, you'll have much more than 2+2 GB1 of RAM and a very good chance everything will stay in RAM anyway.

1 Original matrix + copy.

Branko Dimitrijevic
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  • Actually although this seems like a reasonable suggestion, when running out of memory what happens is _other_ programs get switched out of memory first. The result is the program has a really negative impact on the system because even windows explorer starts to slow down when you get close to the system memory limit. It's better to switch out the data you know you won't use to avoid system slowdown. – bobobobo Jul 01 '12 at 22:56
  • @bobobobo Why would recently-used pages of other programs swap-out before not-recently-used pages of this program? – Branko Dimitrijevic Jul 01 '12 at 23:57
  • I guess because "this" program is active. Relatively speaking, the matrix will have been created recently -- this isn't a background process. – bobobobo Jul 02 '12 at 15:20
  • @bobobobo Virtual memory manager doesn't care much about which memory page belongs to which process. As you copy the matrix, pages will be paged-in back to memory and later will be paged-out again if this or any other process requires more memory. Note that a process doesn't need to be paged-in/out completely - as you continue to work with the copied matrix, OS is free to page-out the original matrix while keeping the copy in memory. – Branko Dimitrijevic Jul 02 '12 at 16:42