Jay Sprenkle wrote: > Here you pass the address of db, which is already a pointer. > >You've passed a pointer to a pointer but never allocated the structure used. > >I think you want to change this: > > >> sqlite3 *db; >> >> >to > > >> sqlite3 db; >> >> Jay,
No, he has the open call correct. He has a local pointer, he passes the address of that pointer to sqlite3_open() and it allocates the sqlite3 structure and sets his pointer to point to it. No need to change this. Derrell has identified his problem. Dennis Cote
