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    <p>Did you actually try this code? It works fine on gcc with a small tweak <code>std::vector&lt;T&gt;::const_iterator</code>, needs to be declared as <code>typename std::vector&lt;T&gt;::const_iterator</code></p> <p>You may be better off with using std::copy and std::ostream_iterator.</p> <p>EDIT: <strong>types, dependent types and typename</strong> Can't fit it all in the comments, so here goes (btw. this is my understanding and I could be off by a country mile - if so please correct me!)...</p> <p>I think this is best explained with a simple example..</p> <p>Let's assume you have a function foo</p> <pre><code>template &lt;typename T&gt; void foo() { T::bob * instofbob; // this is a dependent name (i.e. bob depends on T) }; </code></pre> <p>Looks okay, and typically you may do this</p> <pre><code>class SimpleClass { typedef int bob; }; </code></pre> <p>And call</p> <pre><code>foo&lt;SimpleClass&gt;(); // now we know that foo::instofbob is "int" </code></pre> <p>Again, seems self explanatory, however some nuser comes along and does this</p> <pre><code>class IdiotClass { static int bob; }; </code></pre> <p>Now</p> <pre><code>foo&lt;IdiotClass&gt;(); // oops, </code></pre> <p>What you have now is an expression (multiplication) as IdiotClass::bob resolves to a non-type!</p> <p>To the human, it's obvious that this is stupid, but the compiler has no way of differentiating between types vs. non-types, and by default in C++ (and I think this is where compilers differ), <strong>all</strong> qualified <em>dependent names</em> (i.e. T::bob) will be treated as <em>non-type</em>. To <em>explicitly</em> tell the compiler that the dependent name is a real type, you must specify the <code>typename</code> keyword - </p> <pre><code>template &lt;typename T&gt; void foo() { typedef typename T::bob *instofbob; // now compiler is happy, it knows to interpret "bob" as a type (and will complain otherwise!) }; </code></pre> <p>This applies even if it is a <code>typedef</code>. i.e.</p> <pre><code>template &lt;typename T&gt; void foo() { typedef typename T::bob local_bob; }; </code></pre> <p>Is that any clearer?</p>
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    1. COWhy would `std::copy` be a better way of streaming out in case of a generic vector? With `std::copy` a dedicated long line should be written just to stream out a vector. However with output stream operator it is easier to do: `std::cout << "Assignments = " << assignmentIds << std::endl;`. I would like to print a vector always inside braces `"[" + vector + "]"`, and that would require one more line with just "]" if `std::copy` is used.
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    2. COIt does compile without typename after solving the original problem. Is there any use of specifying `typename` in this context even if it does compile apart from slight risk of having a misinterpreted type?
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    3. CO@Leonid, yes that would be great that someones explain the concept of dependent name.
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