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    <p>Sorry to everybody who wrote "The second half should be similar"... it's not.</p> <p>Anyway, here you go:</p> <pre><code>// traverse array diagonally int c, tmp, x; for (c = N - 1; c &gt; -N; c--) { tmp = N - abs(c) - 1; x = tmp; while (x &gt;= 0) { if (c &gt;= 0) { std::cout &lt;&lt; arr[x][tmp - x] &lt;&lt; ", "; } else { std::cout &lt;&lt; arr[N - (tmp - x) - 1][(N-1)-x] &lt;&lt; ", "; } --x; } std::cout &lt;&lt; "\n"; } </code></pre> <p>Do you need this for a game or something? </p> <p>[edit] looking at this again, I think my answer wasn't very nicely written. Here's a quick run through:</p> <p>Let's pretend that N is 3. </p> <p>What we need is an iteration over coordinate-combinations that looks like this:</p> <pre><code>(0, 0) (1, 0), (0, 1) (2, 0), (1, 1), (0, 2) (2, 1), (1, 2) (2, 2) </code></pre> <p>So first some placeholders:</p> <pre><code>int c, // a counter, set by the outer loop tmp, // for intermediate results x; // the x-index into *arr* (*y* will be defined implicitly) </code></pre> <p>Now this outer loop</p> <pre><code>for (c = N - 1; c &gt; -N; c--) { </code></pre> <p>makes <em>c</em> iterate over <em>{2, 1, 0, -1, 2}</em>. </p> <p>The next step</p> <pre><code> tmp = N - abs(c) - 1; x = tmp; </code></pre> <p>turns <em>{2, 1, 0, -1, -2}</em> into <em>{0, 1, 2, 1, 0}</em>, which are the lengths of the needed outputs at this step minus one (so they can be used as indices). We make two copies of this, <em>tmp</em> and <em>x</em>.</p> <p>Now we count down from <em>x</em> to <em>0</em>:</p> <pre><code> while (x &gt;= 0) { ... --x; } </code></pre> <p>if we're on the upper-left half of <em>arr</em>, indicated by <em>c >= 0</em>, the x-indices into <em>arr</em> need to start at the diagonal and go down to zero <em>(0 to 0, 1 to 0 and 2 to 0)</em> , whereas the y-indices need to start at zero and go up to the diagonal <em>(0 to 0, 0 to 1 and 0 to 2)</em>:</p> <pre><code> if (c &gt;= 0) { std::cout &lt;&lt; arr[x][tmp - x] &lt;&lt; ", "; } </code></pre> <p>once we're on the lower-right half, the x-indices need to start at <em>N</em> and to down to the diagonal <em>(2 to 1 and 2 to 2)</em>, whereas the y-indices need to start at the diagonal and go up to <em>N (1 to 2 and 2 to 2)</em>:</p> <pre><code> else { std::cout &lt;&lt; arr[N - (tmp - x) - 1][(N-1)-x] &lt;&lt; ", "; } </code></pre> <p>finally we just need a line-break at the end of each line:</p> <pre><code> std::cout &lt;&lt; "\n"; </code></pre> <p>Savy? :-)</p>
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    1. COThank you for giving me a complete answer ;). I need this because there is a nested for-loop that cannot be parallelized row-wise or column-wise because of data dependencies, so it can only be done antidiagonal-wise.
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    2. COOk? Sounds scary :-) Well, if it works for you, can you click 'accept' please? Thanks!! :-)
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