Question

当我在c ++中使用一个函数来初始化sqlite3时，当它从函数中出来时，句柄是Null。知道是什么原因引起的吗？我只需将指针作为参数。如果我将open打开到main函数，它可以正常工作。会发生什么导致这个？隐藏的内容是否超出了范围？

#include <iostream>
#include "sqlite3.h"

using namespace std;

int init_table(sqlite3 *dbH,  string db_name)
{
  if (sqlite3_open(db_name.c_str(), &dbH) != SQLITE_OK)
    {
      cout << "Failed to open DB : " << sqlite3_errmsg(dbH) << endl;
      abort();
    }
  else
    {
      cout << "Opened database: " << db_name << endl;
    }

  if (sqlite3_exec(dbH, "PRAGMA synchronous = OFF", NULL, NULL, NULL) != SQLITE_OK)
    {
      cout << "Failed to set synchronous: " << sqlite3_errmsg(dbH) << endl;
    }

  if (sqlite3_exec(dbH, "PRAGMA journal_mode = WAL", NULL, NULL, NULL) != SQLITE_OK)
    {
      cout << "Failed to set journal mode: " << sqlite3_errmsg(dbH) << endl;
    }

  cout << "dbH 2: " << dbH << endl;

}

int main()
{
  sqlite3 * dbH;
  dbH = NULL;
  cout << "dbH 1: " << dbH << endl;
  string dbName = "foo1.db";
  init_table(dbH, dbName);

  cout << "dbH 3: " << dbH << endl;
}

运行时

$ ./a.out
dbH 1: 0
Opened database: foo1.db
dbH 2: 0x5baa048
dbH 3: 0

Answer 1

应该是

int init_table(sqlite3 **dbH,  string db_name)

并将指针传递给指针？
可能是sqliter处理没有问题。它既可以将指针作为引用传递，也可以作为指针传递指针。

当然，在通过时，您需要在修改后将&dbH传递给init_table。

传递给函数调用后，空指针sqlite3句柄

1 个答案: