<p>In <a href="https://github.com/mlpack/mlpack/pull/757#discussion_r74701138">src/mlpack/methods/ann/layer/concat_layer.hpp</a>:</p>
<pre style='color:#555'>&gt; +  {
&gt; +    ForwardTail(layers, output);
&gt; +  }
&gt; +
&gt; +  template&lt;size_t I = 0, typename DataType, typename... Tp&gt;
&gt; +  typename std::enable_if&lt;I == sizeof...(Tp), void&gt;::type
&gt; +  ForwardTail(std::tuple&lt;Tp...&gt;&amp; layers, DataType&amp; output)
&gt; +  {
&gt; +    /* Nothing to do. */
&gt; +  }
&gt; +
&gt; +  template&lt;size_t I = 0, typename DataType, typename... Tp&gt;
&gt; +  typename std::enable_if&lt;I &lt; sizeof...(Tp), void&gt;::type
&gt; +  ForwardTail(std::tuple&lt;Tp...&gt;&amp; layers, DataType&amp; output)
&gt; +  {
&gt; +    output = arma::join_slices(output, std::get&lt;I&gt;(layers).OutputParameter());
</pre>
<blockquote>
<p>but i can write 2 overloads for output type arma::cube and arma::mat. </p>
</blockquote>

<p>I think this is a good solution.</p>

<blockquote>
<p>Is there a better way ?</p>
</blockquote>

<p>I do not know a better way, but it is possible to specialize your function by std::enable_if, you can write something like</p>

<pre><code>//pseudo codes
typename std::enable_if&lt;I &lt; sizeof...(Tp) &amp; std::is_arma_mat&lt;DataType&gt;::value&gt;::type

</code></pre>

<p>Armadillo has a header file define something like std::is_arma_mat, but I forgot where is it. However, this solution is less elegant compare with overload.</p>

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