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copied!<p>You are accessing the value_type trough the variable instance and not the variable type.</p> <p><strong>Method 1</strong> - this works:</p> <pre><code>typedef std::vector<Vector> MyVector; MyVector v; unsigned u = MyVector::value_type::Dim(); </code></pre> <p><strong>Method 2</strong> - or this:</p> <pre><code>std::vector<Vector> v; unsigned u = std::vector<Vector>::value_type::Dim(); </code></pre> <p>If you typedef like on method 1 you do not hardcode assumptions on vector template parameter and you write clean code.</p> <hr> <p><strong>Edit:</strong> Expanded to explain the behavior for this issue as requested by question owner:</p> <p>The scope resolution operator <code>::</code> has higher <a href="http://www.cppreference.com/wiki/operator_precedence" rel="nofollow noreferrer">precedence</a> than any other C++ operator. This includes the Member access from an object <code>.</code> operator. Thus when you write something like:</p> <pre><code>unsigned u= v.value_type::Dim(); </code></pre> <p>this resolves to the following C++ code:</p> <pre><code>unsigned u = v.SVector<double, 2>::Dim(); </code></pre> <p>and ultimately what is resolved first is the <code>SVector<double, 2>::Dim()</code> part. This would force the vector instance declared trough variable <code>v</code> to have a templatized inner class named SVector. And because this does not happen this results in error:</p> <pre><code>error C2039: 'SVector<double,2>' : is not a member of 'std::vector<_Ty>' </code></pre> <p>STL <code>vector</code> would have to be "expanded" for each usage of this pattern (accessing value_type trough variable instance and not variable type). This is not a good solution as it leads to lots of boilerplate and <strong>unnecessary and unmaintainable code</strong>. By following the above mentioned solutions you avoid all this and can easily do what you wanted.</p>

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