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    <p>There is one important reason why it's not enough with the code you proposed:</p> <ul> <li>The assignment operator is applied on a <strong>initialized object</strong>, but this is not the case in the copy constructor. Specifically, your <code>data</code> member is not initialized properly when the code runs <code>*this = array</code> (<code>data</code> isn't in a member-initializer list), so you are calling <code>delete[]</code> operator on the uninitialized <code>data</code>.</li> </ul> <p>It requires you to, at least, use a member-initializer list to initialize the <code>data</code> member before you use it.</p> <p>There are more drawbacks about your propose:</p> <ul> <li>The assignment operator forces restrictions on <code>T</code> type (but maybe that restrictions are in the class <code>Array</code> anyway): <ul> <li><code>T</code> must be default-constructible (as noted by leemes).</li> </ul></li> <li>assignment operator is not good implemented: <ul> <li>is not considering case <code>a = a</code> (self-assignment), as Nick Bougalis noted.</li> </ul></li> <li>is not exception safe: <ul> <li>if <code>new[]</code> operator fails, the object will be in an inconsistent state.</li> <li>if copying <code>T</code> throws an exception, the object will be in an inconsistent state.</li> </ul></li> </ul> <p>I see a much better approach with using the copy constructor in assignment operator implementation instead of the opposite:</p> <pre><code>Array&lt;T, ROW, COL&gt;(const Array&lt;T, ROW, COL&gt; &amp;array): data(new T[ROW * COL]) { for (int i = 0; i &lt; ROW; ++i) { for (int j = 0; j &lt; COL; ++j) { data[j*ROW + i] = other(i, j); } } } Array&lt;T, ROW, COL&gt;&amp; operator=(Array&lt;T, ROW, COL&gt; other) { swap(other); return *this; } void swap(Array&lt;T, ROW, COL&gt;&amp; other) { T* tmp = data; data = other.data; other.data = tmp; } </code></pre> <p>(This is also known as the <strong>copy-and-swap idiom</strong>.)</p>
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    1. POIs this a valid copy constructor?
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    1. POIs this a valid copy constructor?
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    1. COGood first answer, congratulations! May I correct your English?
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    2. COYes, please! Correct it. :)
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    3. CODone! And again, congratulations. You summarized a couple of problems very nicely.
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