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    <p>First, it would be <code>std::vector</code>, and nothing else, because <code>vector</code> resides in the <code>std</code> namespace and you are asking for a <strong>template template parameter</strong>, while <code>std::vector&lt;int&gt;</code> is not a template anymore. Next, a <code>std::vector</code> actually takes <strong>two</strong> template parameters, one for the type and the other for the allocator:</p> <pre><code>template &lt; typename T, template&lt;typename, typename&gt; class ContainerType, typename Alloc = std::allocator&lt;T&gt; &gt; class Stack{ ContainerType&lt;T, Alloc&gt; container; // ... }; // usage: Stack&lt;int, std::vector&gt; s; </code></pre> <p>Now, this only enables containers with two template parameters as the underlying type, so you're better off with what the standard does: take it as a normal type:</p> <pre><code>template &lt;typename T, typename ContainerType&gt; class Stack{ ContainerType container; // ... }; // usage: Stack&lt;int, std::vector&lt;int&gt; &gt; s; </code></pre> <p>To ensure that the underlying type has the same <code>T</code>, you can do a fake "static assert", or if you have a C++0x enabled compiler, you can do an actual static assert:</p> <pre><code>#include &lt;tr1/type_traits&gt; // C++03 us std::tr1::is_same //#include &lt;type_traits&gt; // C++0x, use std::is_same template &lt;typename T, typename ContainerType&gt; class Stack{ typedef typename ContainerType::value_type underlying_value_type; typedef char ERROR_different_value_type[ std::tr1::is_same&lt;T, underlying_value_type&gt;::value ? 1 : -1 ] ContainerType container; // ... }; </code></pre> <p>This works because if <code>T</code> is different from the used container's <code>T</code>, it will be a <code>typedef char ERROR_different_vale_type[-1]</code>, and an array of negative size can't possibly exist, which causes a compiler error. :) Now, with C++0x you can just <code>static_assert</code> that:</p> <pre><code>#include &lt;tr1/type_traits&gt; // C++03 //#include &lt;type_traits&gt; // C++0x template &lt;typename T, typename ContainerType&gt; class Stack{ typedef typename ContainerType::value_type underlying_value_type; static_assert(std::tr1::is_same&lt;T, underlying_value_type&gt;::value, "Error: The type of the stack must be the same as the type of the container"); ContainerType container; // ... }; </code></pre> <p>For convenience, you can now specify a default template argument for the common case:</p> <pre><code>template &lt;typename T, typename ContainerType = std::vector&lt;T&gt;&gt; class Stack{ ContainerType container; // ... }; // usage: Stack&lt;int&gt; s; </code></pre> <p>And at this point you can just use <code>std::stack</code> which does exactly this (though it uses <code>std::deque</code> as the underlying type). :)</p>
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    1. COYes, but I want to be able to write Stack<int, List<int> > stack too and not just vector. If I use your second code block what happens if I use: Stack<int, std::vector<double> > stack. Will it be just a double vector? Can I make a constrain so the vector will be of the type int and not double (I mean so it will be like the first type)?
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    2. CO@user550413: if you write `Stack <int, List>`, then the container type will be `List<int>`, just the same as when `Stack <int, vector>` gets a `vector`.
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    3. CO@user: For the second version, see my edit.
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