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There are countless factors that could be contributing to this.
I would guess that the major ones are file caching and delayed writes[^].
File Caching:
When a file is opened and read, its contents are loaded into RAM by the OS, and then sections are copied to your exe as you need them (usually in 4KB chunks which you then read smaller chunks from). Once the file is closed it is marked as unused but the OS, but is not removed from the RAM. If then another program needs heaps of memory the file will be removed from the RAM, however if this does not happen, and your file is still in the RAM then your program doesn't actually need to use the hard disk.
This is most notable if you open a program that uses lots of files on load, say MS Word. If you then close the program and open it again shortly after without opening something else the 2nd time you open the program it will load much quicker.
Delayed Writes:
These generally only occur to slow mediums such as USB memory sticks, however they can happen to HDD as well.
When you write to a file, and storage device is busy the data you write will often get written into a virtual file in RAM which is then written to the storage device at a later stage.
Other problems may include HDD head seeks (as mentioned by Thaddeus Jones) and other programs accessing the disk.
If you are running Windows vista or 7 then you can look at disk accesses with the Resource Monitor (resmon.exe)
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Thanks Adrew,
I will consider the situation with file caching. Need to investigate on this more.
Meanwhile, are you think that rare disk accesses by other programs can reduce speed of my accesses in 5 times?
I'm making sure that no heavy HDD operations are performed by other programs, but other programs for sure doing disk accesses even in the idle mode.
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If you run out of ideas, have you tried disabling anti-virus software? Maybe it performs some weird caching of scanned data.
Even a long shot is a shot...
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Yes, I tried.
Well, actually I think this is file caching problem. Seems like caching is not always work well in my case. Will try to disable it and do some caching on my own.
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Hi All,
I wanted to know the meaning of a line of code,
a structure is declared in a .h file
struct xyz
{
int q;
char w;
long s;
};
object of it is declared in .cpp file
struct xyz *obj;
obj = ( struct xyz * ) ( buffer + offset );
where buffer is a char array of 1000 bytes and offset is a long variable having 500 value
i want to know meaning of line:-
obj = ( struct xyz * ) ( buffer + offset );
Can anybody help me in this
Thanks in advance
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The structure pointure obj will now point to the memory address where buffer begins, with an offset of 500 added.
Assuming an integer of 32 bits and long of 64 bits, the structure will now have the following data;
obj->q has buffer[500] to buffer[503] as integer value
obj->w is buffer[504]
obj->s has buffer[505] to buffer[512] as long value
modified 13-Sep-18 21:01pm.
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Of course that depends on the current padding. For info see the pack[^] pragma. It could just as well be
obj->q has buffer[500] to buffer[503] as integer value
obj->w is buffer[504]
obj->s has buffer[508] to buffer[515] as long value or something completely different.
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Niklas Lindquist wrote: Of course that depends on the current padding
And, indeed, the size of int and long, as the original response noted. On some 64-bit machines/compiler settings sizeof(int) == 8
Graham
Librarians rule, Ook!
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A bit more explanation: In C/C++ arrays and pointers are almost equivalent (see this C++ Language Tutorial on pointers[^]. Thus buffer + 500 is a pointer that is equivalent to &buffer[500].
This sort of programming construct is often used when reading and writing data to/from devices (such as a disk or a network socket). The device usually works with unstructured data which is treated as a sequence of bytes. Programmers often use char arrays as a buffer (a char is usually the same size as a byte) to read and write to the device. However they wish to use structs within the rest of the program. The cast operator (struct xyz*), instructs the compiler to convert the pointer on the right hand side to a pointer to the xyz struct.
Graham
Librarians rule, Ook!
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I am drawing an image using GDI+ function. Image is having balck color at corners, I need to draw the black color as transparent with window.
How can I do that?
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You can simply use ImageAttributes::SetColorKey function.
ImageAttributes imageAttrs;
Color transparentClr(0, 0, 0);
imageAttrs.SetColorKey(transparentclr, transparentclr);
Use this image attributes in Graphics::DrawImage
graphics.DrawImage(&yourImage, yourRect, 0, 0, yourImage.GetWidth(), yourImage.GteHeight(),
Gdiplus::UnitPixel, &imageAttrs);
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Hi all
I have one Window with size of 700*700.The view size is 500*500 and 100*100 of Opengl.
Now i am drawing some object on 500*500 opengl view.
how i can draw same object on 100 *100 view without create any new object?
means i want to use existing big view's object in Small view.
Let me know how can we archive this thing.
Thanks
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We have the extra large icon view (256 x 256) since Windows Vista.
Now I have a control panel applet running under Win7 and Vista. I have added the icon resources with images of size 256 x 256 into the CPL. But the shortcut (on the Desktop) to this CPL applet does not display the icon correctly.
It is said that assigning the value CPL_DYNAMIC_RES for the idIcon member of CPLINFO structure in the CPL_INQUIRE message handler triggers Windows to send a CPL_NEWINQUIRE message. And then we assign the hIcon member of NEWCPLINFO in the CPL_NEWINQUIRE message handler for the icon information.
But no matter AfxGetApp()->LoadIcon or ::LoadImage (even with the LR_DEFAULTSIZE flag) fails to make the icon image of the shortcut on the Desktop display the correct sized image.
Does anyone know the correct way to handle the CPL_INQUIRE and CPL_NEWINQUIRE to make it display 256 x 256 image for the shortcut on the Desktop?
Thanks in advance.
Maxwell Chen
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The root cause is in the ico file. Do not use PNG (do not compress) for the 256 x 256 and 128 x 128 frames. Use BMP (uncompressed) instead. And the issue is resolved.
Maxwell Chen
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I have some structure like
struct abc{
int x;
char y;
DWORD z;
__int64 a;
char data[250];
}
I have this structure in some tree and i want to save tree contents in a file. Later i want read this file and fill the tree
is there any library available for that?
or is there any simple and fast way to do this
please sugget
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In C you'd use fread and fwrite and in C++ you'd use std::istream::read and std::ostream::write to save plain old data (POD) structures without pointers to a file. All the functions I've mentioned read and write raw bytes so if you want a human readable file you'll have to use something else e.g. use formatted I/O in C and the stream insertion/extraction operators in C++.
Cheers,
Ash
modified on Monday, January 31, 2011 4:38 AM
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simplest way are fwrite and fread.
WriteFile and ReadFile APIS are another options.
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Hi,
I am drawing an image and moving it using mouse move. It is working fine. But If I load 2 MB image it is taking time to draw and also taking time to get refresh every time when I move the image.
How can I reduce the flicking?
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instead of loading the image every time, load it first time, save in some memory dc or buffer. BitBilt this to your screen dc. You may use the whole drawings on another memory dc first, then copy it to window dc to avoid flickering.
see double buffering
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Hi, I am using GDI+ function to darw image.
Graphics *g2 = new Graphics(pDc->GetSafeHdc());
g2->DrawImage(...);
How could I use double buffering concept here? Please give any example.
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Bitmap membitmap(width, height, PixelFormat32bppPARGB);
Graphics memgraphics(&membitmap);
Graphics *g2 = new Graphics(pDc->GetSafeHdc());
g2->DrawImage(&membitmap);
An eg. here
Also, you can use CachedBitmap to optimize the rendering if you are loading and drawing same bitmap again and again.
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Hi John,
You can solve your problem by implementing double buffering concept...ie you can do all the drawings in a bitmap and once your drawings are over just bitblt it in to your client DC..
Regards,
spk521
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