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    copied!<p>You say you need a pointer to an iterator... but that's quite confusing and rarely needed.</p> <p>Let's see your solutions, one at a time:</p> <p><strong>Solution 1:</strong></p> <pre><code>vector&lt;type&gt;::iterator *it = &amp;vector&lt;type&gt;::_First; </code></pre> <p>Not valid because <code>_First</code> is private, and also you have to use the <code>vector_object</code>, not the class name.</p> <p><strong>Solution 2:</strong></p> <pre><code>vector&lt;type&gt;::iterator *it = &amp;(vector&lt;type&gt;::begin()); </code></pre> <p>Not valid because <code>begin()</code> is not a static method, so you have to use the object name, not the class name:</p> <pre><code>vector&lt;type&gt;::iterator *it = &amp;(vector_object.begin()); </code></pre> <p>But even then, not valid because <code>begin</code> returns an r-value, and you cannot get the address of such an expression. You have to use an auxiliary variable for that, which will be an l-value.</p> <pre><code>vector&lt;type&gt;::iterator *it = &amp;&amp;(vector&lt;type&gt;[0]); </code></pre> <p>Same as above about the staticness of member functions. It should be:</p> <pre><code>vector&lt;type&gt;::iterator *it = &amp;&amp;(vector_object[0]); </code></pre> <p>But even then, to be fair, this code <em>is</em> crazy. <code>&amp;&amp;</code> is the logical AND operator. You cannot apply operator <code>&amp;</code> (address-of) twice in a row because it has to be applied to an l-values but it returns an r-value. Moreover <code>operator[]</code> returns a value of <code>type</code> not an iterator.</p> <p><strong>Solution 3:</strong></p> <p>This compiles but renders undefined behavior because you are returning the address of a local variable that is destroyed once you return from the function. Returning the address of a local variable is always an error, and most compilers will issue a warning about it.</p> <p>What you have to do is to keep the variable around as long as you need the pointer:</p> <pre><code>vector&lt;type&gt;::iterator it = vector_object.begin(); vector&lt;type&gt;::iterator *pit = &amp;it; // use pit, but keep it around </code></pre> <p>That is, your pointer is valid only as long as <code>it</code> variable exists. That's not something specific to iterators, but to pointers in general.</p> <p>The big question here is: why on earth do you need a pointer to an iterator? I've been programming STL for a very long time and I have never needed a pointer to an iterator before (pointers-to-pointers yes, even pointers-to-pointers-to-pointers, but pointers-to-iterators... never).</p>
 

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