title: ImageView src and background tags when using <layer-list> with a centered <bitmap>
body: <p>I have an <code>ImageView</code>, and a <code>&lt;layer-list&gt;</code> in <code>/res/drawable/</code>. This is the layer-list:</p>
  
  <pre><code>&lt;layer-list xmlns:android="http://schemas.android.com/apk/res/android" &gt;
  &lt;item&gt;
      &lt;!-- This is a 24dp x 24dp PNG --&gt;
      &lt;bitmap android:src="@drawable/ic_secondary_arrow" 
              android:gravity="center"/&gt;
  &lt;/item&gt;
  &lt;item&gt;
      &lt;shape android:shape="oval"&gt;
          &lt;stroke android:color="@color/icon_primary_background"
                  android:width="2dp"/&gt;
          &lt;solid android:color="@android:color/transparent" /&gt;
      &lt;/shape&gt;
  &lt;/item&gt;
  &lt;/layer-list&gt;
  </code></pre>
  
  <p>ImageView has nothing special:</p>
  
  <pre><code>&lt;ImageView
      android:id="@+id/arrow"
      android:layout_width="40dp"
      android:layout_height="40dp"
      android:layout_margin="16dp"/&gt;
  </code></pre>
  
  <p>When using <code>src</code> or <code>setImageDrawable(int)</code> pointing to the layer-list, <code>center</code> seems to be ignored (or at least this is not my desired behavior):</p>
  
  <p><img src="https://i.stack.imgur.com/8dnte.png" alt="Undesired behavior"></p>
  
  <p>I want to display it as below. This is possible using <code>background</code> (or <code>setBackground(Drawable)</code>) instead of <code>src</code>:</p>
  
  <p><img src="https://i.stack.imgur.com/oLZRF.png" alt="Desired behavior"></p>
  
  <p>The reason I insist is because <code>setBackground(Drawable)</code> is limited to API16+, and I'd rather not limit my audience because of this.</p>
  
  <p>So, what is it that I'm missing?</p>
  
  <p><strong>Edit:</strong> as @pskink pointed below, I was missing <code>setBackgroudDrawable(Drawable)</code>. But still, that was an incidental question. I really want to know what's the reason behind the different behavior of <code>src</code> and <code>background</code> (and the respective methods) in this case.</p>