Allowing User To Select Directory..

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12-13-2006
kwikness

Allowing User To Select Directory..

In this function, I'm trying to allow the user to select which directory the mp3 playlist is saved to..

Code:

void savePlaylist() { string directoryToSaveTo; cout << "What directory would you like to load a playlist from?" << endl << "The file must be a file named playlist.wpl" << endl << "Example: C:\\" << endl; cin >> directoryToSaveTo; cout << endl << endl; directoryToSaveTo = directoryToSaveTo + "playlist.wpl"; ofstream playlist; playlist.open(directoryToSaveTo); //compiler error here, the open function is //supposed to be used like this: //playlist.open("c:\playlist.wpl"); playlist << "<?wpl version=\"1.0\"?>" << endl << "<smil>" << endl << " <head>" << endl << " <meta name=\"Generator\" content=\"Microsoft Windows Media Player -- 10.0.0.4036\"/>" << endl << " <author/>" << endl << " <title>Playlist</title>" << endl << " </head>" << endl << " <body>" << endl << " <seq>" << endl; for(int j = 1; fileAddresses[j] != ""; j++) playlist << " <media src=\"" << fileAddresses[j] << "\"/>" << endl; playlist << " </seq>" << endl << " </body>" << endl << "</smil>" << endl; playlist.close(); cout << "\nPlaylist saved to c:\\playlist.wpl" << endl; quitMessage(); }

I want to be able to have the user select where the playlist gets saved, but I get the compiler error:

Quote:

cannot convert parameter 1 from 'std::string' to 'const wchar_t *'

What can I do to get around this problem?
12-13-2006
Oldman47

You're better of with char;

Code:

int main() { char directoryToSaveTo[255]; std::cout << "What directory would you like to load a playlist from?" << endl << "The file must be a file named playlist.wpl" << endl << "Example: C:\\" << endl; std::cin >> directoryToSaveTo; std::cout << endl << endl; strcat (directoryToSaveTo,"playlist.wpl"); ofstream playlist; playlist.open(directoryToSaveTo); //no compiler error here ;) }
12-13-2006
Mario F.

the open() method of the playlist object expects const wchar_t*. A pointer to a const wchar_t, which on this case almost certainly refers to a const array of wchar_t.

The string object cannot convert automatically to an array of wchar_t.
I'm not used to wide character arrays. However I think it will be enough if you pass it directoryToSaveTo.str(), which will pass a C-Style string and I reckon the implicit conversion to a wchar_t array will happen.
12-13-2006
vart

it is called c_str()
12-13-2006
Mario F.

c_str(), yes. Mark that.
12-13-2006
Salem

> You're better of with char;
Sticking with std::string and learning how to use it properly is better IMO.
Rushing back to unsafe C char arrays isn't the way to go.
12-13-2006
kwikness

Thanks guys. Used c_str() and it worked fine.
12-13-2006
CornedBee

Where did wchar_t make it into the mix? Even wide streams take narrow-char filenames.
12-13-2006
Mario F.

from this:

Quote:

cannot convert parameter 1 from 'std::string' to 'const wchar_t *'
12-13-2006
CornedBee

Yes, but how did the compiler come up with this message? There is no open(const wchar_t*) member function in basic_ifstream.
12-13-2006
Mario F.

It happens when playlist.open() gets called with a std::string parameter. My guess is that this is someone else's library.
12-13-2006
dwks

Maybe open() is overloaded to accept wchar_t* and char*.
12-14-2006
CornedBee

Hmm, apparently there are two things working together.
First, the MS standard library provides wchar_t overloads of open().
Second, the MS compiler produces really stupid error messages. (It should offer all overloads if it can't call any.)

I see ...