The Blob Example [1] contains code [2] in which, if sqlite3_prepare() fails, the subsequent call to sqlite3_finalize() is skipped. Is this OK? Does sqlite3_prepare() free up any memory it may have allocated if it fails?
I've read the documentation for these two functions but still don't know. Thanks, Jerry Krinock [1] http://www.sqlite.org/cvstrac/wiki?p=BlobExample [2]: static int writeBlob( sqlite3 *db, const char *zKey, const unsigned char *zBlob, int nBlob) { const char *zSql = "INSERT INTO blobs(key, value) VALUES(?, ?)"; sqlite3_stmt *pStmt; int rc; do { rc = sqlite3_prepare(db, zSql, -1, &pStmt, 0); if( rc!=SQLITE_OK ){ return rc; } sqlite3_bind_text(pStmt, 1, zKey, -1, SQLITE_STATIC); sqlite3_bind_blob(pStmt, 2, zBlob, nBlob, SQLITE_STATIC); rc = sqlite3_step(pStmt); assert( rc!=SQLITE_ROW ); rc = sqlite3_finalize(pStmt); } while( rc==SQLITE_SCHEMA ); return rc; } _______________________________________________ sqlite-users mailing list sqlite-users@sqlite.org http://sqlite.org:8080/cgi-bin/mailman/listinfo/sqlite-users
