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copied!<blockquote> <pre><code> value=cvQueryFrame( capture ); </code></pre> <p>when I change value, lastValue changes too</p> </blockquote> <p>No, it doesn't. The pointer <code>value</code> gets overwritten, as you desire. It is impossible for this line of code to affect <code>lastValue</code>.</p> <blockquote> <p>but because they are pointers, both end up with the same value</p> </blockquote> <p>No, it doesn't matter that they are pointers. Pointers are still objects in and of themselves.</p> <hr> <p><strong>However</strong>, <code>cvQueryFrame</code> returns a pointer to a buffer which <a href="http://www.cs.iit.edu/~agam/cs512/lect-notes/opencv-intro/opencv-intro.html" rel="nofollow">you shall not modify or free</a> as this is done for you:</p> <blockquote> <p>Note that the image captured by the device is allocated/released by the capture function. There is no need to release it explicitly. </p> </blockquote> <p>Though <a href="http://opencv.jp/opencv-1.0.0_org/docs/ref/opencvref_highgui.htm" rel="nofollow">the documentation</a> is a little unclear, it seems likely to me that the buffer is only valid until the next call to <code>cvQueryFrame</code> (which will then re-use the allocated memory). So even though <code>lastValue</code> can't and doesn't change, it <em>happens</em> to end up pointing to the new frame anyway.</p> <p>To get around this, you can explicitly copy the object that <code>lastValue</code> points to:</p> <pre><code>lastValue = cvCloneImage(value); </code></pre> <p><em>Now</em> you probably take on responsibility for freeing it (but again it's not entirely clear from my cursory glance at the documentation):</p> <pre><code>cvReleaseImage(&lastValue); </code></pre>

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