title: Two UIScrollView in one UIView
body: <p>I am trying to add two <code>UIScrollView</code>'s into one <code>UIView</code>, Both the <code>scrollView</code>'s show up properly, the problem that I am having is how to determine which <code>scrollView</code> is scrolled because based on that I have to populate the images. This is what I am doing:</p>
  
  <p>I have a <code>ViewController</code> with <code>UIScrollViewDelegate</code>.</p>
  
  <p>In the <code>loadView</code> method of my <code>ViewConroller</code>, I do the following:</p>
  
  <pre><code>CGRect scrollViewFrame1;
  CGPoint scrollViewPoint1;
  scrollViewPoint1.x = 0;
  scrollViewPoint1.y = 57;
  
  CGSize scrollViewSize1;
  scrollViewSize1.width = 320;
  scrollViewSize1.height = 154;
  
  scrollViewFrame1.size = scrollViewSize1;
  scrollViewFrame1.origin = scrollViewPoint1;
  
  CGRect scrollViewFrame2;
  CGPoint scrollViewPoint2;
  scrollViewPoint2.x = 0;
  scrollViewPoint2.y = 258;
  
  CGSize scrollViewSize2;
  scrollViewSize2.width = 320;
  scrollViewSize2.height = 154;
  
  scrollViewFrame2.size = scrollViewSize2;
  scrollViewFrame2.origin = scrollViewPoint2;
  
  scrollView1 = [[UIScrollView alloc] initWithFrame:scrollViewFrame1];
  scrollView2 = [[UIScrollView alloc] initWithFrame:scrollViewFrame2];
  </code></pre>
  
  <p>And then: </p>
  
  <pre><code>  scrollView1.delegate = self;
    scrollView2.delegate = self;
  </code></pre>
  
  <p>And then: </p>
  
  <pre><code> [self.view addSubView:scrollView1];
   [self.view addSubView:scrollView2];
  </code></pre>
  
  <p>I have one <code>scrollViewDidScroll:</code> method, how do I determine which <code>scrollView</code> this method got invoked by, because based on that i need to populate different images for my <code>scrollView</code>.</p>
  
  <pre><code> - (void)scrollViewDidScroll:(UIScrollView *)scrollView
  {
          //Code to populate the next or previous images for scrollView
              // If it was one i am able to show the images
  }
  </code></pre>
  
  <p>Thanks for the help.</p>