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If you read and write data from the same disk from different files, the read-head will have to move to a different place on the disk, which is slow. If your input and output file are on different drives, this problem is solved. If you read and write alternatively (to the same physical disk) in your program each time, this slow down is maximum.
Note that some variation in disk speed can't be prevented in a multitasking OS where other programs might be using the disk too.
If your memory permits, you could consider reading the entire input file to memory at once at the beginning of your algorithm and then processing it and writing the results.
modified 13-Sep-18 21:01pm.
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My files are always one the same drive in all tests. When it does in 1:54 or 24 seconds.
Thaddeus Jones wrote: Note that some variation in disk speed can't be prevented in a multitasking OS where other programs might be using the disk too.
I always ensure no other work is running. Also I'm looking in the "Resource monitor", no big HDD accesses except my program.
Thaddeus Jones wrote: If your memory permits, you could consider reading the entire input file to memory at once at the beginning of your algorithm and then processing it and writing the results.
No this is not possible, algorithm should work with unlimited file size.
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Maybe you could increase your input buffer then from 512 bytes to say 100Mb, and every time you've processed the 100Mb from memory, you'll read a new 100Mb. Similarly, writing your output to a memory buffer (say also 100Mb) and once your buffer is full writing that to file, should help with speed too.
The idea is to concentrate disk access to areas on the disk that are near eachother, since those access operations are much faster than if the head has to be repositioned.
modified 13-Sep-18 21:01pm.
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Thaddeus Jones wrote: Maybe you could increase your input buffer then from 512 bytes to say 100Mb, and every time you've processed the 100Mb from memory, you'll read a new 100Mb. Similarly, writing your output to a memory buffer (say also 100Mb) and once your buffer is full writing that to file, should help with speed too.
I've tried this approach. Although it gives slight speed increase, it doesnt resolve the problem of chaotic speed.
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I'm afraid I'm out of ideas then 
modified 13-Sep-18 21:01pm.
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The problem is not "Algorithm is slow", but "Algorithm speed is too chaotic". Sometimes it's done 5 times faster then usual, this means - it should always be done 5 times faster.
I agree, that access speed can vary a little, but 5 times... I think this is something that should be puzzled out.
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progDes wrote: Usually this algorithm finishes in 1 minute and 50 seconds. But I noticed that sometimes (rarely) it finishes in 24 seconds! Processing the same file, the same execution path. I've started to search for the reasons of why this slowdown is happening and how can I control that.
Caching, perhaps?
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"Man who follows car will be exhausted." - Confucius
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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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