如果有一段WPF程序：

XAML:

<Window x:Class="WpfApplication1.Window1"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        Title="Window1" Height="300" Width="300">
    <Grid>
        <Button Content="Button" HorizontalAlignment="Left" Margin="34,30,0,0" VerticalAlignment="Top" Width="75" Click="Button_Click_1" x:Name="buton1"/>
        <TextBlock HorizontalAlignment="Left" Margin="58,90,0,0" TextWrapping="Wrap" Text="TextBlock" VerticalAlignment="Top" x:Name="textBlock1"/>
    </Grid>
</Window>

Code Behind:

        private void Button_Click_1(object sender, RoutedEventArgs e)
        {
            this.textBlock1.Text = "hello world!";
        }

逻辑很简单，这段程序在点击Button之后TextBlock会显示“hellow world!”，但是如果我把程序用Task变一下，然它变成异步模式执行：

        private void Button_Click_1(object sender, RoutedEventArgs e)
        {
            Task.Factory.StartNew(() => this.textBlock1.Text = "hello world!");
        }

就会出现跨线程操作异常，因为
Task.Factory.StartNew(Action a) 会创建一个Task，这个Task运行在另外一个不同于UI的线程上。解决方法也很简单，就是我提到的用Dispatcher.BeginInvoke

        private void Button_Click_1(object sender, RoutedEventArgs e)
        {
            Task.Factory.StartNew(() => 
                this.buton1.Dispatcher.BeginInvoke(new Action(() => this.textBlock1.Text = "hello world!")));
        }

That is..