Question

我编写了2个简单函数。

第一个与我的数据库建立连接并启用外键支持：

<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/twitter-bootstrap/3.3.5/css/bootstrap.min.css">

<link rel="stylesheet" href="nv.1.8.1.d3.css">

第二个关闭连接：

int hmd_db_connect (sqlite3 *db) {
    char *sql;
    sqlite3_stmt *stmt;

    if (sqlite3_open_v2(HMD_DB_PATH, &db,
            SQLITE_OPEN_FULLMUTEX | SQLITE_OPEN_SHAREDCACHE,
            NULL) != SQLITE_OK) {
        hmd_log_err("Failed to connect to database [Path: %s]: %s.",
                HMD_DB_PATH, sqlite3_errmsg(db));
        return -1;
    }

    sql = hmd_make_str("PRAGMA foreign_keys = ON");

    if (sqlite3_prepare_v2(db, sql, strlen(sql), &stmt, NULL) != SQLITE_OK) {
        hmd_log_err("Failed to prepare SQL [%s]: %s.",
                sql, sqlite3_errmsg(db));
        hmd_db_disconnect(db);
        free(sql);
        return -1;
    }

    if (sqlite3_step(stmt) != SQLITE_DONE) {
        hmd_log_err("Failed to enable foreign key support: %s.",
                sqlite3_errmsg(db));
        sqlite3_finalize(stmt);
        hmd_db_disconnect(db);
        free(sql);
        return -1;
    }

    sqlite3_finalize(stmt);
    free(sql);
    return 0;
}

然后在我的int hmd_db_disconnect (sqlite3 *db) { if (sqlite3_close(db) != SQLITE_OK) { hmd_log_err("Failed to disconnect from database [Path: %s]: %s.", HMD_DB_PATH, sqlite3_errmsg(db)); return -1; } return 0; }函数中，我将这两个函数称为：

main

程序没有显示任何错误消息。但是当我用_info("Connecting to database.\n"); hmd_db_connect(db); _info("Database connected.\n"); _info("Disconnecting from database.\n"); hmd_db_disconnect(db); _info("Database disconnected.\n");运行它时，它显示可能的内存泄漏：

valgrind

我已经检查了==11079== LEAK SUMMARY: ==11079== definitely lost: 0 bytes in 0 blocks ==11079== indirectly lost: 0 bytes in 0 blocks ==11079== possibly lost: 57,768 bytes in 47 blocks ==11079== still reachable: 0 bytes in 0 blocks ==11079== suppressed: 0 bytes in 0 blocks ==11079== ==11079== For counts of detected and suppressed errors, rerun with: -v ==11079== ERROR SUMMARY: 47 errors from 47 contexts (suppressed: 6 from 6)的日志，在sqlite3的共享库中检测到所有错误，并且我没有看到任何不自由的变量。

请问好吗？

Answer 1

我认为sqlite3*不正确。它应该返回sqlite3*，但需要sqlite3**而不是sqlite3*。您只需丢失sqlite3_open_v2返回的sqlite3* db = NULL; hmd_db_connect(db);，想象一下：

db

即使hmd_db_connect成功打开数据库，

int hmd_db_connect(sqlite3 **db) { char *sql; sqlite3_stmt *stmt; if (sqlite3_open_v2(HMD_DB_PATH, db, SQLITE_OPEN_FULLMUTEX | SQLITE_OPEN_SHAREDCACHE, NULL) != SQLITE_OK) { hmd_log_err("Failed to connect to database [Path: %s]: %s.", HMD_DB_PATH, sqlite3_errmsg(*db)); return -1; } sql = hmd_make_str("PRAGMA foreign_keys = ON"); if (sqlite3_prepare_v2(*db, sql, strlen(sql), &stmt, NULL) != SQLITE_OK) { hmd_log_err("Failed to prepare SQL [%s]: %s.", sql, sqlite3_errmsg(*db)); hmd_db_disconnect(*db); free(sql); return -1; } if (sqlite3_step(stmt) != SQLITE_DONE) { hmd_log_err("Failed to enable foreign key support: %s.", sqlite3_errmsg(*db)); sqlite3_finalize(stmt); hmd_db_disconnect(*db); free(sql); return -1; } sqlite3_finalize(stmt); free(sql); return 0; }仍然为NULL。

我修改了代码：

<a onClick="sessionStorage.someKey = 'category 1'; console.log(sessionStorage.someKey);" data-fajax="false" href="some-page.html" data-transition="slidefade">Category 1</a>

现在似乎没有泄漏。

sqlite3的共享库中可能存在内存泄漏

1 个答案: