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    <p>With your allocation pattern, you can't do a simple copy because of the multiple allocations in the second loop. It depends on what you're really after. You can arrange things so that you do just one allocation of a set of pointers — by setting up pointers into your current 1D array so that you can access it as a 2D array. Alternatively, you can create a copy of your 1D array and then set up the pointers (2 allocations, 1 copy, 1 initialization loop). Since the only issue is whether to copy the original allocation, the main answer will work on the original data.</p> <p>You have:</p> <pre><code>int N, A, B; A = ...; B = ...; N = A * B; double *a_1d = malloc(N * sizeof(double)); ...check and load a_1d... </code></pre> <p>Now you can create a 2D 'array' by:</p> <pre><code>double **a_2d = malloc(A * sizeof(double *)); ...check that allocation succeeded... for (int i = 0; i &lt; A; i++) a_2d[i] = &amp;a_1d[i * B]; </code></pre> <p>With this initialization done, you can now use:</p> <pre><code>for (int i = 0; i &lt; A; i++) for (int j = 0; j &lt; B; j++) a_2d[i][j] = ...; </code></pre> <p>If you really need to duplicate the original <code>a_1d</code> array, you'd add:</p> <pre><code>double *a_1d_copy = malloc(N * sizeof(double)); ...check that the allocation succeeded... memmove(a_1d_copy, a_1d, N * sizeof(double)); </code></pre> <p>and then you can reference <code>a_1d_copy</code> instead of <code>a_1d</code> in setting up <code>a_2d</code>.</p> <p>Remember that when it comes time to free things, you need to release both <code>a_1d</code> and <code>a_2d</code> (and <code>a_2d_copy</code> if you're using that).</p>
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    1. POcan I copy a 1d array into a 2d structure without using a loop in c?
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    1. USJonathan Leffler
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