Background: How caching works

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You know that? The longer you work with your Linux system, the bigger the memory consumption is. However, performance is not negatively affected. Looking closely at it, you see only the cache has grown bigger. The article writes a program that eats up all memory, then terminates with a std::bad_alloc. Before terminating, it has of course eaten up all your cache and transformed into usr memory. Then, the memory is freed and there is no more cache.

main.cpp
#include <iostream>
using namespace std;

void pollute()
{
  int* i=new int();
  cout << i << "  ";
}

int main()
{
  while (true) { pollute(); }
}
Compile, link and run it:
g++ main.cpp && ./a.out

How can you use this

Imagine, you are doing a file system read benchmark. Your system is fresh:

tweedleburg:~ # free
             total       used       free     shared    buffers     cached
Mem:       4053216     795664    3257552          0        352      54624
-/+ buffers/cache:     740688    3312528
Swap:            0          0          0

You have 54624 bytes in all caches.

tweedleburg:~ # dd if=wine-1.0-rc2.tar of=/dev/null
197360+0 records in
197360+0 records out
101048320 bytes (101 MB) copied, 2.278 s, 44.4 MB/s

You get 44.4 MB/s for disk reads, a realistic result.

tweedleburg:~ # dd if=wine-1.0-rc2.tar of=/dev/null
197360+0 records in
197360+0 records out
101048320 bytes (101 MB) copied, 0.190445 s, 531 MB/s

At the second time, you get 531 MB/s for disk reads, an unrealistically good result. The culprit are the caches that stored the file content:

tweedleburg:~ # free
             total       used       free     shared    buffers     cached
Mem:       4053216     886360    3166856          0        528     145748
-/+ buffers/cache:     740084    3313132
Swap:            0          0          0

You use my program to clear the caches:

tweedleburg:~ # ./a.out >/dev/null
terminate called after throwing an instance of 'St9bad_alloc'
  what():  std::bad_alloc
Aborted

My program allocates memory till it can no more. All caches are eaten up. Then it stops and frees its memory:

tweedleburg:~ # free
             total       used       free     shared    buffers     cached
Mem:       4053216     794916    3258300          0        344      57360
-/+ buffers/cache:     737212    3316004
Swap:            0          0          0

You repeat the file read:

tweedleburg:~ # dd if=wine-1.0-rc2.tar of=/dev/null
197360+0 records in
197360+0 records out
101048320 bytes (101 MB) copied, 2.21617 s, 45.6 MB/s

and get a realistic result. To show you I am not telling bullshit, we repeat the disk read without cleaning the caches and get again an unrealistic result:

tweedleburg:~ # dd if=wine-1.0-rc2.tar of=/dev/null
197360+0 records in
197360+0 records out
101048320 bytes (101 MB) copied, 0.165862 s, 609 MB/s    

You see there is a big difference between a cached and a non-cached read; but also there is a difference to a direct read:

tweedleburg:~ # dd iflag=direct if=wine-1.0-rc2.tar of=/dev/null
197360+0 records in
197360+0 records out
101048320 bytes (101 MB) copied, 23.6831 s, 4.3 MB/s

reading with dd using the direct flag yields the same performance as reading with an empty cache, you just have to adjust the blocksize:

tweedleburg:~ # dd iflag=direct if=wine-1.0-rc2.tar bs=1024k of=/dev/null
96+1 records in
96+1 records out
101048320 bytes (101 MB) copied, 2.42799 s, 41.6 MB/s