Question

I’m writing a light wrapper for the SQLite API.

Basically, I’m curious about how/when an SQLite pre-compiled statement gets executed…

When i go:

char buffer[] = "INSERT INTO example VALUES (?1, ?2, ?3, ?4, ?5, ?6, ?7)";
sqlite3_stmt* stmt;
sqlite3_prepare_v2(mDb, buffer, strlen(buffer), &stmt, NULL);

for (unsigned i = 0; i < mVal; i++)
{
    std::string id = getID();
    sqlite3_bind_text(stmt, 1, id.c_str(), id.size(), SQLITE_STATIC);
    sqlite3_bind_double(stmt, 2, getDouble());
    sqlite3_bind_double(stmt, 3, getDouble());
    sqlite3_bind_double(stmt, 4, getDouble());
    sqlite3_bind_int(stmt, 5, getInt());
    sqlite3_bind_int(stmt, 6, getInt());
    sqlite3_bind_int(stmt, 7, getInt());

    if (sqlite3_step(stmt) != SQLITE_DONE)
    {
        printf("Commit Failed!\n");
    }

    sqlite3_reset(stmt);
}

sqlite3_finalize(stmt);

At what point is the actual sql executed? Is it during the call to sqlite3_prepare_v2, or during the first sqlite3_step?

Any clarity is greatly appreciated 🙂

Cheers

Jarrett

Answer 1

According to the SQLite documentation for sqlite3_prepare, we see:

To execute an SQL query, it must first be compiled into a byte-code program using
one of these routines:
sqlite3_prepare, sqlite3_prepare_v2, sqlite3_prepare16, sqlite3_prepare16_v2.

And for sqlite3_step:

After a prepared statement has been prepared using either sqlite3_prepare_v2() or
sqlite3_prepare16_v2() or one of the legacy interfaces sqlite3_prepare() or
sqlite3_prepare16(), this function must be called one or more times to evaluate
the statement.

More information:

SQLite has a virtual machine that does all necessary actions to execute your code on the selected database. sqlite3_prepare (and its family) compiles your SQL statement into byte code that can be executed on this virtual machine. On the other hand, sqlite3_step executes that byte-code in the VM.