how can I applied to my code?

I mean I have the scan drives

I can use getdirs() == DRIVE_FIXED

I can use getdirs() == DRIVE_FIXED

that's one of these issues. No you cannot, since getdirs() returns void. You should probably have another function e.g. writeToDrive that takes the drive letter as parameter and you call it in your case DRIVE_FIXED: block

I tried to changed to BOOL or string , but didn’t work

I tried to changed to BOOL or string , but didn’t work

That requires your function to return something. But that's the other issue. Your function can return only once. So it's "hard" to keep track of which drives you already processed, the best way would be to call a function in the case DRIVE_FIXED: block

then should I iterate szDrive and writE?

these freakin’ winapi names … szKek, WIN32_SHOULD_PWORD_KEK_LOL_P macros

why not something like this? (might not compile, due to wide-chars, just to show the idea) void writeToDrive(char drive) { std::string filename = drive + ":\\file.txt"; std::ofstream myfile; myfile.open (filename); // do your thing myfile.close() } // ... case DRIVE_FIXED: writeToDrive(szDrive[0]); break; // ...

why drive[0] if I am passing a simple vari without array?

why drive[0] if I am passing a simple vari without array?

you can also pass the char and not the pointer

E0167 argument of type "WCHAR" is incompatible with parameter of type "char *"

I got this

https://gist.github.com/SkyBulk/3a809fc2dbe393b60827ddaef33daf39

E0167 argument of type "WCHAR" is incompatible with parameter of type "char *"

oh yeah right.. try #include <cstdlib> void writeToDrive(WCHAR wdrive) { char drive; std::wctomb(&drive, wdrive);

Error C2664 'void writeToDrive(WCHAR)': cannot convert argument 1 from 'WCHAR [5]' to 'WCHAR' strix

Error C2664 'void writeToDrive(WCHAR)': cannot convert argument 1 from 'WCHAR [5]' to 'WCHAR' strix

how do you pass the parameter / call the function? like this writeToDrive(szDrive[0]); ?

I forgot since I changed to non array version

writeToDrive(szDrive[0]);

it didnt write nothing

I have this drives

https://gist.github.com/SkyBulk/3a809fc2dbe393b60827ddaef33daf39

https://gist.github.com/SkyBulk/3a809fc2dbe393b60827ddaef33daf39

You can try this, this also tells you when a file cannot be opened / written. Writing to C:\ might require higher rights void writeToDrive(WCHAR wdrive) { std::string filename{"_:\\file.txt"}; std::wctomb(&filename[0], wdrive); std::ofstream file(filename); if (!file) { std::printf("Error, could not open file %s\n", filename.c_str()); } else { file << "<<<<<<<<<<<<<<<<<<<< Heya!>>>>>>>>>>>>>>>>>>>>\n"; file.close(); } }

it could write on D: , but not on C:

permissions issues?

permissions issues?

yes - Writing to C:\ might require higher rights

i have group nt

why not c?

self-protection?

thanks a lot @olli

self-protection?

Yes, by default the user has no right to write to C. This also happens if you try to copy file. sure, no worries :)

if i am nt group then should I switch from user to nt and do the process?

Can anyone help me with linux library linking

Can anyone help me with linux library linking

#meta

#ot pls

hi guys! why " std::endl " is used ?

Flushes the buffer

Flushes the buffer

yeah it is written the same on the book that i am learning

but i need more explanation

When you do cout it collects data before putting it to your console

what does " flushing the buffer " mean

When you do cout it collects data before putting it to your console

And when flushing it actually “sends” that buffer to a console and clears the collected data

so it is used for memory management, right?

to make the program lighter?

to make the program lighter?

Buffered IO is used quite often to improve efficiency. Since IO operations (e.g. writing to the screen) are quite expensive the content is written into a buffer first. Once the buffer is full or you flush it, the content will actually be transmitted (e.g. displayed).

hi guys! why " std::endl " is used ?

std::cout, by default, doesn't print anything unless you flush its buffer. std::endl prints a new line and flush the buffer, so the text is printe. It's equivalent to "\n" << std::flush;

to make the program lighter?

No, << std::endl is actually slower compared to << '\n' and is equivalent to << '\n' << std::flush.

so it is to print a new line and clear the temporarily stored data

??

am i right?

Yes

To clear the buffer

+ a new line

if i dont use it the collected data will remain in my program

if i dont use it the collected data will remain in my program

Collected data?

Collected data?

printed words for example

The text that you streamed to std::cout might not be printed if your program is killed unexpectedly

Imagine having a big string where you append all the text that goes to std::cout and everytime you flush that string is cleared and your terminal prints the text that was in it

And it's important to mention that a flush is quite expensive, so you should avoid flushing in a loop for example

so it is to print a new line and clear the temporarily stored data

It guarantees your content will be written immediately. It has no effect on the data in your application.

And it's important to mention that a flush is quite expensive, so you should avoid flushing in a loop for example

what do you mean by expensive?

what do you mean by expensive?

Slow

uhuh

thanks

if i dont use it the collected data will remain in my program

If you flush after writing to the file, you will se an update to the file every 5 seconds, oftherwise it might take until the end for the file to be changed #include <thread> #include <chrono> #include <fstream> int main() { std::ofstream fp("file.txt"); for (int i = 0; i < 10; ++i) { fp << i; // fp.flush(); // add or remove this line std::this_thread::sleep_for( std::chrono::seconds{5} ); } fp.close(); // content will be flushed here // if not flushed already return 0; }

btw Olli, do you know if there's something in the standard that specifically prevents std::cout << "Hello World!" << '\n'; from being as fast as std::cout << "Hello World!\n"; Like if the compiler sees that there are two subsequent streams whose parameters are of the same type is there anything that prevents the optimizer from simply merging them?

I know it doesn't do it, so that's why I asked if it is a limitation of the standard

Or just the optimizer not being smart enough

Speaking of which it is actually calling different operator<<'s here

Sorry for phone picture, but I don't have a PC rn

I've just done some benchmarking and it doesn't merge those calls

hey guys it is again me

is <chrono> standard library?

is <chrono> standard library?

Yep

I've just done some benchmarking and it doesn't merge those calls

Yeah it doesn't, but I'm curious if it could or the standard says that they should be separate

Because if they aren't noexcept, then technically one could throw an exception and merging there would change behavior

Lemme check if they are

Yep

why i am getting this when i try to include <chrono>