title: scala macro how to convert `HList` to function args
body: <p>For below types</p>
  
  <pre><code>type HFunc = (Int :: String :: HNil) =&gt; Int
  
  type Func = (Int, String) =&gt; Int
  </code></pre>
  
  <p>I try to convert <code>Func</code> to <code>HFunc</code></p>
  
  <pre><code>val funExpr: Tree = ???
  val hlistType = ???      
  val hfuncName = c.freshName("hfunc")
  
  q"""
    def $hfuncName(t: $hlistType) = {
      ${funExpr}(..) //how to extract hlist elements as params ?
    }
  """
  </code></pre>
  
  <p>How Can I extract the <code>HList</code> elements and pass it to the <code>Func</code> ?</p>