SQLite creates DB/file instead of opening the existing one

I am trying to open an SQLite database in python, using the sqlite shared library (libsqlite3.0.dylib on macOS). (I use this library instead of the sqlite3 module, because I need to use the sqlite3_update_hook function, which is not exposed in the python module.)

The organization tree is the following:

MyProject/
    SRC/
        - ProductionSQLite.py
        - CONST.py
        DATA/
            - imdb-tiny.db 
            - Some bigger .db files

The file called "imdb-tiny.db" is an SQLite database.
The file CONST.py only contains constants for other scripts.

I am trying to open this database from the python script, using the following code:

from ctypes import CDLL, byref
from sys import stderr

DATA_PATH = "DATA/"
DLL_NAME = "libsqlite3.0.dylib" # {"Linux": "libsqlite3.so", "MacOS": "libsqlite3.0.dylib", "Windows": "winsqlite3.dll"}

SQLITE3 = CDLL(DLL_NAME) # Load sqlite3 shared library

if __name__=="__main__":

	# Pointeur vers la DB (plus précisément, vers un objet qui la représente)
	db = c_void_p()
	# Liaison avec la base SQLite
	return_number:int = SQLITE3.sqlite3_open(f"{DATA_PATH}imdb-tiny.db", byref(db))
	if return_number != 0:
		# Erreur à l'ouverture de la DB
		print(f"Echec de l'ouverture de la DB: erreur {return_number}.", file=stderr)
		print(f"Documentation des erreurs: https://www.sqlite.org/capi3ref.html#SQLITE_ABORT", file=stderr)
		exit(-1)

	# Exécution de requêtes sur la base SQLite
	query = b'CREATE TABLE foo (id INT, name VARCHAR(255))'
	err = c_char_p() # Create a variable to hold error messages
	SQLITE3.sqlite3_exec(db, query, None, None, byref(err))
	if err: print(err.value, file=stderr)

The scripts runs without any error, but it does not behave as expected. It creates a new file/database, called with the first letter of the database name (here, it would be "D" as it is the first char of the string "DATA/imdb-tiny.db"). The location of the created file depends on weather I'm executing from VSCode extension (MyProject folder), or a terminal (same location as the terminal). Any idea on why this is happening and how to fix it ?

The problem here is that you don't encode the string into a list of utf-8 bytes for the sqlite3_open function call. The D is the first character of the string.

By explicitly encoding the string into a byte list, you'll be able to open the correct file:

SQLITE3.sqlite3_open(f"{DATA_PATH}imdb-tiny.db".encode('utf-8'), byref(db))

The underlying problem is that python has no hints as to the data types that are passed as parameters to the function, but regardless, you need to ensure that the parameters are binary encoded in the correct format for use by the function.