title: How a non async value is wrapped in a Future using await?
body: <p><a href="https://dart.dev/guides/language/language-tour#handling-futures" rel="nofollow noreferrer">Docs</a> says:</p>
  <blockquote>
  <p>In await expression, the value of expression is usually a Future; if it isn’t, then the value is automatically wrapped in a Future.</p>
  </blockquote>
  <p>So, if I use:</p>
  <pre class="lang-dart prettyprint-override"><code>void main() async {
    var a = await 1;
    print(a); // Prints 1
    print(Future.value(1)); // Prints `Instance of _Future&lt;int&gt;`. 
  }
  </code></pre>
  <p>Using what the docs says, the first print should print <code>Instance of _Future&lt;int&gt;</code> but it prints <code>1</code>. Can anyone explain what that line means by <code>the value is automatically wrapped in a Future</code>?</p>