-event- 只能出现在 += 或 -= 的左侧

发布于 2024-10-09 00:34:11 字数 466 浏览 0 评论 0原文

我有一个循环事件。我试图防止同一方法多次添加到一个事件中。我已经实现了 add 和 remove 访问器。

但是，我收到一条错误消息：

ItemsProcessed 只能出现在 += 或 -= 的左侧

当我尝试调用它们时，即使在同一个类中也是如此。

ItemsProcessed(this, new EventArgs()); // Produces error

public event EventHandler ItemsProcessed
{
    add
    {
        ItemsProcessed -= value;
        ItemsProcessed += value;
    }
    remove
    {
        ItemsProcessed -= value;
    }
}

原文

I have an event in a loop. I am trying to prevent the same method being added to an event more than once. I've implemented the add and remove accessors.

However, I get an error stating that:

ItemsProcessed can only appear on the left hand side of += or -=

When I try to call them, even within the same class.

ItemsProcessed(this, new EventArgs()); // Produces error

public event EventHandler ItemsProcessed
{
    add
    {
        ItemsProcessed -= value;
        ItemsProcessed += value;
    }
    remove
    {
        ItemsProcessed -= value;
    }
}

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帅气称霸 2024-10-16 00:34:11

对于显式事件，您需要提供自己的后备存储 - 委托字段或类似 EventHandlerList 的内容。当前的代码是递归的。尝试：

private EventHandler itemsProcessed;
public event EventHandler ItemsProcessed
{
    add
    {
        itemsProcessed-= value;
        itemsProcessed+= value;
    }

    remove
    {
        itemsProcessed-= value;
    }
}

然后（并注意到我对“即将变为 null”边缘情况重新线程持谨慎态度）：

var snapshot = itemsProcessed;
if(snapshot != null) snapshot(this, EventArgs.Empty);

使用更新的 C# 版本，这可以被简化：

itemsProcessed?.Invoke(this, EventArgs.Empty);

With an explicit event, you need to provide your own backing store - either a delegate field or something like EventHandlerList. The current code is recursive. Try:

private EventHandler itemsProcessed;
public event EventHandler ItemsProcessed
{
    add
    {
        itemsProcessed-= value;
        itemsProcessed+= value;
    }

    remove
    {
        itemsProcessed-= value;
    }
}

Then (and noting I'm being a little cautious about the "about to turn null" edge-case re threading):

var snapshot = itemsProcessed;
if(snapshot != null) snapshot(this, EventArgs.Empty);

With more recent C# versions, this can be simplified:

itemsProcessed?.Invoke(this, EventArgs.Empty);

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逆流 2024-10-16 00:34:11

我无法从您的帖子中判断您是否试图从派生类引发事件，但我发现的一件事是您无法在基类中定义事件，然后（直接）引发它在派生类中，出于某种原因，我还不太清楚。

因此，我在基类中定义受保护的函数来引发事件（在这些基类中定义），如下所示：

// The signature for a handler of the ProgressStarted event.
// title: The title/label for a progress dialog/bar.
// total: The max progress value.
public delegate void ProgressStartedType(string title, int total);

// Raised when progress on a potentially long running process is started.
public event ProgressStartedType ProgressStarted;

// Used from derived classes to raise ProgressStarted.
protected void RaiseProgressStarted(string title, int total) {
    if (ProgressStarted != null) ProgressStarted(title, total);
}

然后在派生类中，我调用 RaiseProgressStarted(title,total) 而不是调用 ProgressStarted(title,total)。

看起来好像绕了很远的路。也许其他人知道解决这个问题的更好方法。

I can't tell from your post if you are trying to raise the event from a derived class or not, but one thing I've found is that you can't define an event in a base class and then raise it (directly) in a derived class, for some reason that isn't real clear to me yet.

So I define protected functions in base classes to raise events (that are defined in those base classes), like this:

// The signature for a handler of the ProgressStarted event.
// title: The title/label for a progress dialog/bar.
// total: The max progress value.
public delegate void ProgressStartedType(string title, int total);

// Raised when progress on a potentially long running process is started.
public event ProgressStartedType ProgressStarted;

// Used from derived classes to raise ProgressStarted.
protected void RaiseProgressStarted(string title, int total) {
    if (ProgressStarted != null) ProgressStarted(title, total);
}

Then in the derived class, I call RaiseProgressStarted(title, total) instead of calling ProgressStarted(title, total).

It seems like kind of the long way around. Maybe someone else knows of a better way around this problem.

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