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copied!<p><code>N</code> does not get "initialized" to anything. It is not a variable. It is not an object. <code>N</code> is a compile-time constant. <code>N</code> only exists during compilation. The value of <code>N</code> as well as the actual <code>T</code> is determined by the process called <em>template argument deduction</em>. Both <code>T</code> and <code>N</code> are deduced from the actual type of the argument you pass to your template function. </p> <p>In the first call the argument type is <code>int[6]</code>, so the compiler deduces that <code>T == int</code> and <code>N == 6</code>, generates a separate function for that and calls it. Let's name it <code>cal_size_int_6</code></p> <pre><code>void cal_size_int_6(int (&a)[6]) { std::cout << "size of array is: " << 6 << std::endl; } </code></pre> <p>Note that there's no <code>T</code> and no <code>N</code> in this function anymore. Both were replaced by their actual deduced values at compile time.</p> <p>In the first call the argument type is <code>int[1]</code>, so the compiler deduces that <code>T == int</code> and <code>N == 1</code>, generates a separate function for that as well and calls it. Let's name it <code>cal_size_int_1</code></p> <pre><code>void cal_size_int_1(int (&a)[1]) { std::cout << "size of array is: " << 1 << std::endl; } </code></pre> <p>Same thing here. </p> <p>Your <code>main</code> essentially translates into</p> <pre><code>int main() { int a[]={1,2,3,4,5,6}; int b[]={1}; cal_size_int_6(a); cal_size_int_1(b); } </code></pre> <p>In other words, your <code>cal_size</code> template gives birth to <em>two different functions</em> (so called <em>specializations</em> of the original template), each with different values of <code>N</code> (and <code>T</code>) hardcoded into the body. That's how templates work in C++.</p>

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