c – 获取用户输入时的分段错误

我有这个：

// Get database access parameters 
    const char* db = "codes",*server = "localhost",*user = "root",*pass = "pass";

    // Connect to the sample database.
    MysqLpp::Connection conn(false);
    if (conn.connect(db,server,user,pass)) {

        for (int i=0; i<10; ++i)
        {
            int d,count;
            cout << "Введите D\n";
            cin >> d;
            cout << "Введите количество записей при D=" << d << endl;
            cin >> count;
            for (int a=0; a<count; ++a)
            {
                char * name;
                int r,n1,n2;
                cout << "Введите R\n";
                cin >> r;
                cout <<"Введите n1 и n2\n";
                cin >> n1 >> n2;
                cout <<"Введите название\n";
                cin >> name;
                MysqLpp::Query query = conn.query();
                for (int j=n2-n1+1; j<n2; ++j)
                {
                    int k =pow(2,(j+r));
                    query << "insert into code (n,k,d,name) values (" << j << "," << k << "," <<d<<"," << MysqLpp::quote_only << name << ");";
                    query.execute(); 
                }
            }
        }
        conn.disconnect ();
        return 0;
    }
    else {
        cerr << "DB connection Failed: " << conn.error() << endl;
        return 1;
    }
}

很抱歉发布了所有代码.
我可以编译程序没有错误,但在命令行中,它说“分段错误”.这个错误是什么？以及如何解决？

解决方法

你应该分配空间

char * name;

在写信之前.

如果您知道该名称不会更长,那么MAXNAME则只需定义名称即可

char name[MAXNAME+1]

那应该可以解决你的问题.