I'm trying to list the logical drives of my Pc and add the result to a vector but I'm getting this weird result = ;;;.
What am i missing here?
std::vector<std::string> directory::getLogicalDrives() { DWORD mydrives = 100; char lpBuffer[100]; DWORD drives = GetLogicalDriveStrings(mydrives, lpBuffer); std::vector<std::string> driveList; for (int i = 0; i < 100; i++) { std::string drive(3, '%c' + lpBuffer[0]); // Missing something? driveList.push_back(drive); } return driveList; } As stated in the documentation GetLogicalDriveStrings() gives you a list of NULL-terminated strings, and the list is terminated by a NULL character. So just iterate that list, eg:
std::vector<std::string> directory::getLogicalDrives() { std::vector<std::string> driveList; char szBuffer[105]; DWORD size = GetLogicalDriveStrings(104, szBuffer); if ((size > 0) && (size <= 104)) { const char *pstr = szBuffer; while( *pstr ) { std::string drive( pstr ); driveList.push_back(drive); pstr += (drive.length() + 1); } } return driveList; } 
GetLogicalDriveStringsW, replace char with wchar_t, and std::string with std::wstring.
DWORD mydrives = 100; char lpBuffer[100];
100 is not enough characters. Technically, a computer can have up to 26 drive letters (realistically, no one has that many drive letters at one time, but you should still prepare for it). A 100-character buffer has enough space to receive 24 drive letter strings at most (4 * 24 + 1 = 97, 4 * 25 + 1 = 101). You need space for at least 105 characters in your buffer to receive 26 drive letter strings (4 * 26 + 1 = 105).
std::string drive(3, '%c' + lpBuffer[0]); // Missing something?
This line does not do what you think it does. You are thinking of the C-style sprint() function instead. You can't use formatting strings with std::string like this. This line should not even compile, or should compile with warnings.
Besides, you are not even looping through the output strings correctly anyway. GetLogicalDriveStrings() returns a double-terminated list of strings in <drive>:\<nul> format, where <drive> is <nul> at the end of the list. For example, if drives A, B, and C are returned, the contents of the buffer will look like this:
lpBuffer[ 0] = 'A'; lpBuffer[ 1] = ':'; lpBuffer[ 2] = '\\'; lpBuffer[ 3] = '\0'; lpBuffer[ 4] = 'B'; lpBuffer[ 5] = ':'; lpBuffer[ 6] = '\\'; lpBuffer[ 7] = '\0'; lpBuffer[ 8] = 'C'; lpBuffer[ 9] = ':'; lpBuffer[10] = '\\'; lpBuffer[11] = '\0'; lpBuffer[12] = '\0'; The correct approach is to loop through the buffer in groups of 4 characters until you reach that final null terminator (and don't forget error checking!), eg:
DWORD drives = GetLogicalDriveStrings(mydrives, lpBuffer); if ((drives > 0) && (drives <= mydrives)) { char *pdrive = lpBuffer; while (*pdrive) { std::string drive(pdrive); driveList.push_back(drive); pdrive += 4; } } Alternatively:
DWORD drives = GetLogicalDriveStrings(mydrives, lpBuffer); if ((drives > 0) && (drives <= mydrives)) { for (int i = 0; i < drives; i += 4) { std::string drive(&lpBuffer[i]); driveList.push_back(drive); } } That being said, consider using GetLogicalDrives() instead, then you don't have to worry about allocating memory for, or looping through, any drive letter strings at all:
DWORD drives = GetLogicalDrives(); if (drives) { char drive[] = "?:\\"; for (int i = 0; i < 26; i++) { if (drives & (1 << i)) { drive[0] = 'A' + i; driveList.push_back(drive); } } } 
lpBuffer[0]when'%c' + lpBuffer[0]resolves to;?