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    <p>Think of it this way, suppose you had a bunch of functions:</p> <pre><code>// Note that you don't need to name the array, since you don't // actually reference the parameter at all. size_t array_size(const int (&amp;)[1]) { return 1; } size_t array_size(const int (&amp;)[2]) { return 2; } size_t array_size(const int (&amp;)[3]) { return 3; } // etc... </code></pre> <p>Now when you call this, which function gets called?</p> <pre><code>int a[2]; array_size(a); </code></pre> <p>Now if you templatize the arraysize, you get:</p> <pre><code>template &lt;int n&gt; size_t array_size(const int (&amp;)[n]) { return n; } </code></pre> <p>The compiler will attempt to instantiate a version of array_size that matches whatever parameter you call it with. So if you call it with an array of 10 ints, it will instantiate array_size with n=10. </p> <p>Next, just templatize the type, so you can call it with more than just int arrays:</p> <pre><code>template &lt;typename T, int n&gt; size_t array_size(const T (&amp;)[n]) { return n; } </code></pre> <p>And you're done.</p> <p><strong>Edit</strong>: A note about the <code>(&amp;)</code></p> <p>The parentheses are needed around the <code>&amp;</code> to differentiate between array of int references (illegal) and reference to array of ints (what you want). Since the precedence of <code>[]</code> is higher than <code>&amp;</code>, if you have the declaration:</p> <pre><code>const int &amp;a[1]; </code></pre> <p>because of operator precedence, you end up with a one-element array of const references to int. If you want the <code>&amp;</code> applied first, you need to force that with parentheses:</p> <pre><code>const int (&amp;a)[1]; </code></pre> <p>Now the you have a const reference to a one element array of ints. In the function parameter list, you don't need to specify the name of a parameter if you don't use it, so you can drop the name, but keep the parentheses:</p> <pre><code>size_t array_size(const int (&amp;)[1]) </code></pre>
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    1. POCan someone explain this template code that gives me the size of an array?
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    1. COHow come you need the parentheses around the & in the template version of the function, but not in the examples above where the array size is hard-coded?
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    2. COYou actually need the parentheses in both situations.
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