[英]How to work with the 4th argument of the sqlite3_exec function?
我想將一個std::list
傳遞給我的回調 function。我知道我必須在我的回調 function 中使用void*
,但是在使用void*
* 時我該如何將對象插入到我的列表中?
//function will insert to a list of albums (data) a new album
int callbackAlbums(void* data, int argc, char** argv, char** azColName)
{
Album album;
for(int i = 0; i < argc; i++)
{
if (std::string(azColName[i])) == "NAME")
{
album.setName(std::string(argv[i]));
}
else if (std::string(azColName[i])) == "CREATION_DATE")
{
album.setCreationDate(std::string(argv[i]));
}
else if (std::string(azColName[i])) == "USER_ID")
{
album.setOwner(atoi(argv[i]));
}
}
data.push_back(album);
return 0;
}
//function will return a list of all albums.
const std::list<album> DatabaseAccess::getAlbums()
{
const char* sqlStatement = "SELECT * FROM ALBUMS;";
std::list<Album> albums;
char* errMessage = "";
int res = sqlite3_exec(this->_db, sqlStatement, callbackAlbums, (void*)&albums, &errMessage);
}
我嘗試將回調 function 中的void*
更改為std::list<Album>
並期望sqlite3_exec()
與該回調 function 一起工作。
您已經知道如何將list*
轉換為void*
(順便說一句,顯式轉換是多余的)。 只需在回調中做相反的轉換,例如:
int callbackAlbums(void* data, int argc, char** argv, char** azColName)
{
Album album;
...
std::list<Album>* albums = (std::list<Album>*)data;
albums->push_back(album);
...
}
並且不要忘記從getAlbums()
實際return
您的std::list
:
const std::list<album> DatabaseAccess::getAlbums()
{
std::list<Album> albums;
...
int res = sqlite3_exec(..., callbackAlbums, &albums, ...);
...
return albums;
}
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