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    <p>Could you try this?</p> <pre><code>Range rAll = Range::all() ; Mat *rgbMat = new Mat(bgraMat, &amp;rAll); </code></pre> <p>EDIT: PK. Good that it solved the problem. Just to help you undertand why I proposed this:</p> <p>You want to use the constructor:</p> <pre><code> Mat::Mat(const Mat&amp; m, const Range* ranges) </code></pre> <p>It take a reference to a <code>Mat</code> variable. You simple pass one <code>Mat</code> variable. And take a pointer to a <code>Range</code> variable. You can pass the adress of one <code>Range</code> variable. <code>Range</code> just store two <code>int</code>:<code>start</code> and <code>end</code>. It is copiable. One special "static" vale of <code>Range</code> is what <code>all()</code> return and what you can copy and pass in <code>rAll</code>. You will need to check that this work rigth and that not just compile. </p> <p>EDIT 2: Trying to understand the meaning of what are you doing, I will propose simple:</p> <pre><code>Mat *rgbMat = new Mat(bgraMat); </code></pre> <p>or</p> <pre><code>Mat rgbMat = Mat(bgraMat); </code></pre> <p>I think it will be the same as:</p> <pre><code>Range rAll[2]; rAll[0]= Range::all() ; rAll[1]= Range::all() ; Mat *rgbMat = new Mat(bgraMat, rAll); </code></pre> <p>as @Peter Wood pinted. Also I'm curios about:</p> <pre><code>long lrgba = (long) mRgba; Mat bgraMat = *(Mat*) lrgba; </code></pre> <p>If mRgba is a pointer to the Mat you want to clone, why not?:</p> <pre><code>Mat *rgbMat = new Mat(*mRgba); </code></pre>
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    1. COThe docs say `ranges` has a range for each dimension of `Mat m`. In your case if `m` isn't one dimensional, you'll likely have problems.
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    2. COShould be `Mat *rgbMat = new Mat(bgraMat, rAll);`, not `&rAll`.
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    3. CO@Peter Wood . Thank!
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