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    <p>What's biting you here is C's (and C++'s) lack of by-reference argument passing. The <code>a</code> in <code>main()</code> is distinct from the <code>a</code> in <code>func()</code>:</p> <ol> <li><code>a</code> is declared in <code>main</code>.</li> <li><code>a</code> is passed by value (because there's no other way) to <code>func</code>.</li> <li>The <code>a</code> in <code>func</code> is assigned to.</li> <li><code>func</code> returns. Its <code>a</code> is destroyed (leaking memory), and <code>main</code>'s <code>a</code> is left uninitialised.</li> <li>???</li> <li><code>main</code> tries to use <code>a</code>. <strong><em>Segfault!</em></strong></li> </ol> <p>There are a few possible solutions here:</p> <ol> <li><p>Do it the classic C way: pass a pointer to the value. In this case, the parameter will be <code>int ***a</code>, which is gettng a little ridiculous, but whatever.</p> <pre><code>void func(int ***a, int x, int y) { *a = new int*[x]; for (int i=0; i&lt;x; i++) (*a)[i] = new int[y]; } int main(int argc, char **argv) { ... int **a; func(&amp;a, 2, 3); ... } </code></pre></li> <li><p>Do it the C++ way: pass a reference.</p> <pre><code>void func(int **&amp;a, int x, int y) { // ^ // This is the only change. Neat! </code></pre></li> <li><p>Do it the proper (in my opinion) way: return a value from the function, and initialise the array from that.</p> <pre><code>int **func(int x, int y) { int **a = new int*[x]; // EDIT: Fixed stupid typo bug // existing code return a; } int main(int argc, char **argv) { ... int **a = func(2, 3); ... } </code></pre></li> </ol>
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    1. POInitialize Array of Pointers by Reference C++
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    1. COThe second way was what i was looking for, now it works flawlessly. Thanks for comparing the 3 different ways.
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