Re: Why is network reading slow?

"Tom Serface" <>
Thu, 14 Jan 2010 22:38:34 -0800

My experience has been that if the buffer gets too large it starts to slow
down the operation. In my case, I have to read all sizes of files and I've
found the optimal buffer to be around 16K (I think that's what OP was

I use the SDK functions CreateFile, ReadFile, WriteFile, rather than MFC and
my copy routine is around 2-3x what the DOS copy command does (or CopyFile).
I had to write my own for a special purpose though because I have to glue
files back together that span multiple volumes, but I was happy to get
better performance than I was getting with CopyFile previously.

I think there is a trade-off somewhere, but I'm not entirely sure where. I
just did trial and error with different scenarios until I got it the best I


"Joseph M. Newcomer" <> wrote in message

See below...
On Thu, 14 Jan 2010 09:01:47 -0600, "Peter Olcott" <>

"Hector Santos" <> wrote in message

Peter Olcott wrote:

"Hector Santos" <> wrote in
message news:%23OQCOfNlKHA.1824@TK2MSFTNGP04.phx.gbl...

Peter Olcott wrote:

By File Copy, you mean DOS copy command or the
CopyFile() API?

I am using the DOS command prompt's copy command. This
is fast.

The problem is the contradiction formed by the fact that
reading and writng the file is fast, while reading and
not wrting this same file is slow.
I am currently using fopen() and fread(); I am using
Windows XP.

True, if the DOS copy command is fast,then I believe the
code you are using is not optimal. The DOS Copy is using
the same CreateFile() API which fopen() also finally uses
in the RTL. So you should be able to match the same
performance of the DOS Copy command.

Have you tried using setvbuf to set a buffer cache?

Here is a small test code that opens a 50 meg file:

// File: V:\wc7beta\testbufsize.cpp
// Compile with: cl testbufsize.cpp

#include <stdio.h>
#include <windows.h>

void main(char argc, char *argv[])
   char _cache[1024*16] = {0}; // 16K cache
   BYTE buf[1024*1] = {0}; // 1K buffer

Reading a 50MB file, why such an incredibly tiny buffer?

   FILE *fv = fopen("largefile.dat","rb");
   if (fv) {
       int res = setvbuf(fv, _cache, _IOFBF,
       DWORD nTotal = 0;
       DWORD nDisks = 0;
       DWORD nLoops = 0;
       DWORD nStart = GetTickCount();
       while (!feof(fv)) {

The memset is silly. Wastes time, accomplishes nothing. You are setting
a buffer to 0
right before completely overwriting it! This is like writing
int a;

a = 0; // make sure a is 0 before assigning b
a = b;

            int nRead = fread(buf,1,sizeof(buf),fv);
            nTotal +=nRead;
            if (nRead > 0 && !fv->_cnt) nDisks++;
       printf("Time: %d | Size: %d | Reads: %d | Disks:

If I were reading a small 50MB file, I would do

void tmain(int argc, _TCHAR * argv[])
   HANDLE h = CreateFile(_T("largefile.dat"), GENERIC_READ, 0, NULL,


   GetFileSizeEx(h, &size);

   // This code assumes file is < 4.2GB!
   LPVOID p = VirtualAlloc(NULL, (SIZE_T)size.LowPart, MEM_COMMIT,
   DWORD bytesRead;
   ReadFile(h, p, size.LowPart, &bytesRead, NULL);
   ... process data
   VirtualFree(p, (SIZE_T)size.LowPart, MEM_DECOMMIT);
   return 0;

Note that the above does not do any error checking; the obvious error
checking is left as
an Exercise For The Reader. No read loops, no gratuitous memsets, just
simple code that
does exactly ONE ReadFile.

What this basically shows is the number of disk hits it
by checking the fv->_cnt value. It shows that as long as
the cache size is larger than the read buffer size, you
get the same number of disk hits. I also spit out the
milliseconds. Subsequent runs, of course, is faster since
the OS API CreateFile() is used by the RTL in buffer mode.

Also do you know what protocol you have Samba using?

I am guessing that the code above will work with a file of
any size?
If that is the case, then you solved my problem.
The only Samba protocol that I am aware of is smb.


Joseph M. Newcomer [MVP]
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