Question

我正在尝试从静态方法初始化静态变量，但在初始化之后它仍然是null。我做错了什么？

class Database {
    public static Database Connection = null;

    public static void Create() {
        Database.Connection = new Database();

        if (Database.Connection == null) {
            Console.WriteLine("Null");
        }
    }

    public Database() {
        Console.WriteLine("I got called");
    }
}

我在这里遗漏了什么吗？调用方法后，Database.Connection为NULL，尽管已经调用了构造函数。

Answer 1

class Database {
    public static Database Connection = null;

    static Database() {
        Database.Connection = new Database();

        if (Database.Connection == null) {
            Console.WriteLine("Null");
        }
    }

    public Database() {
        Console.WriteLine("I got called");
    }
}

你永远不会打电话给Create()。如果您愿意，可以使用静态构造函数，如上所述。

Answer 2

使用Static constructor初始化静态字段。

static Database()
{
    Database.Connection = new Database();
}

Answer 3

一个选项是使静态成员成为属性：

class Database
{
    private static Database _connection = null;
    public static Database Connection
    {
        get
        {
            if(null == _connection)
            {
                _connection = new Database();
            }
            return _connection;
        }
    }
}

如果您正在实现单例，那么您应该通过将其设为私有来隐藏构造函数。

private Database()
{
    // This will only be called once, when the Connection
    // property getter is first accessed and the private 
    // _connection field is still null.  Note: when using 
    // the debugger, this may be before it is actually called
    // explicitly from within your code!
    Console.WriteLine("Database() constructor called");
}

初始化变量为空？

3 个答案: