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copied!<p>You only need to <code>delete</code> memory you allocate with <code>new</code>.</p> <blockquote> <p>printf would print 5.0, so the variable pointed to by a is not exactly destroyed.</p> </blockquote> <p>You're actually running into undefined behavior. Although the value is still there, the memory was released and can be reused.</p> <p>So the following:</p> <pre><code>struct CoordLocation { float X; float Y; float Z; }; </code></pre> <p>can't create a memory leak if you omit the destructor.</p> <p>Your next snippet:</p> <pre><code>struct CoordLocation { float *X; float *Y; float *Z; }; int main() { CoordLocation *coord = new CoordLocation(); delete coord; return 0; } </code></pre> <p>can potentially create a memory leak, but not as it is. The following will:</p> <pre><code>int main() { CoordLocation *coord = new CoordLocation(); coord->X = new float(); delete coord; return 0; } </code></pre> <p>Your third example</p> <pre><code>struct CoordLocation { CoordLocation() { *X = new float; *Y = new float; *Z = new float; } ~CoordLocation() { delete X; delete Y; delete Z; } float *X; float *Y; float *Z; }; int main() { CoordLocation *coord = new CoordLocation(); delete coord; return 0; } </code></pre> <p>won't create a memory leak because you free all the memory that you allocate. If you were to omit the destructor or forget to call <code>delete coord;</code>, they you'd have a memory leak.</p> <p>A good rule of thumb: call a <code>delete</code> for every <code>new</code> and a <code>delete[]</code> for every <code>new[]</code> and you're safe.</p>

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