title: Passing an "empty" argument to a SASS mixin so that the default argument gets used
body: <p>Consider the following two mixins:</p>
  
  <pre><code>@mixin padding($v: 8px)
    padding: $v - 3 $v $v - 3 $v
  
  @mixin button($padding: "")
    @include padding($padding)
    @include border-radius
    border: 1px solid black
  </code></pre>
  
  <p>I can call <code>@include padding</code> without an argument, and the default <code>8px</code> gets used, no problem. But how can I call the <code>button</code> mixin and still have the default value get used in the <code>padding</code> mixin? If I simply call <code>@include button</code> without an argument, the <code>button</code> mixin still seems to pass some kind of argument to <code>padding</code> (which <code>padding</code> inevitably interprets as 0, thus not adding any padding). Based on <a href="https://stackoverflow.com/questions/3525007/making-a-sass-mixin-with-optional-arguments">Making a Sass mixin with optional arguments</a>, I tried modifying the <code>button</code> mixin to have</p>
  
  <p><code>@include padding(unquote($padding))</code></p>
  
  <p>but that had no effect -- <code>padding</code> still seemed to interpret the argument as being 0, rather than there being no argument and thus using its default value.</p>