Need an explanation of how pointers work when passing in as function args

c

Solution

Most likely, `sqlite3_open` is allocating memory for the database handle. For this reason the function needs a pointer to a pointer to the database handle (`sqlite3`) so that it can modify the pointer to the database handle. For example:

typedef struct { /*...*/ } sqlite3;

int sqlite3_open(const char *filename, sqlite3 **ppDb) {
    /* ... */

    // Allocate memory for the database handle.
    *ppDb = (sqlite3 *)malloc(sizeof(sqlite3));

    /* ... */
    return 0;
}

However, `sqlite3_close` only needs a single pointer to `free` the memory:

int sqlite3_close(sqlite3 *pDb) {
    /* ... Cleanup stuff ... */

    free(pDb);

    return 0;
}

Problem

I thought I understood the basics of pointers, but after checking out some documentation on some sqlite3 methods I got thrown, so now I am unsure if my understanding is correct. Here is a call to an sqlite3 method: ``` char* dataFilePath = "foobar.sqlite"; if (sqlite3_open(dataFilePath, &database) != SQLITE_OK) {...} ``` And here is the function header declaration: ``` int sqlite3_open( const char *filename, /* Database filename (UTF-8) */ sqlite3 **ppDb /* OUT: SQLite db handle */ ); ``` Why is it that &database suddenly becomes a pointer to a pointer? Another method call to close the database connection is: sqlite3_close(database); With the following at the function header: ``` int sqlite3_close(sqlite3 *); ``` Why is this just a pointer, when I pass in a pointer? Would this not be a pointer to a pointer? From all examples I have seen it always seemed the inverse of the functions above, ie. ``` // function void foo(someDataType *bar) { ... } // function call foo(&bar); ``` Thanks for the help.

Original source