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    <p>This statement has several problems:</p> <pre><code>p=(int (*(*p)[b])[a])malloc(b*sizeof(int (*p)[a])); </code></pre> <p>First, <code>malloc</code> returns a <code>void*</code>. You are casting that pointer using <code>(int (*(*p)[b])[a])</code> which yields a value, not a data type. That isn't a valid cast, so that's one reason that the compiler is yelling at you. At this point, <code>p</code> hasn't been initialized so the de-referencing taking place here can crash your program if this statement was executed.</p> <p>Inside your <code>malloc</code> call, you are using <code>sizeof(int (*p)[a])</code>. The statement <code>int (*p)[a]</code> isn't a valid C statement.</p> <p>It seems that you are making this a bit more complex that it needs to be. There are two ways of building a 2D array. You can build an array using <code>malloc(a * b * sizeof(int))</code> as Reinderien explains. You can also build a 1D array of pointers, each pointing to an array of type <code>int</code>. From your code, it seems you are trying to do the latter.</p> <p>The easier way to do this would be something like this:</p> <pre><code>int **p; ... get input from user ... // Declare an array of int pointers of length b p = malloc(b * sizeof(int*)); // For each int* in 'p' ... for (i = 0; i &lt; b; ++i) { // ... allocate an int array of length 'a' and store a pointer in 'p[i]' .. p[i] = malloc(a * sizeof(int)); // ... and fill in that array using data from the user printf("\t\bEnter Column %d\t\n"); for(j = 0; j &lt; a; j++) scanf("%d", &amp;p[i][j]); } </code></pre> <p>Using this method of building a 2D array allows you to use the syntax <code>p[x][y]</code>. Since <code>p</code> is a pointer-to-pointer, <code>p[x]</code> is a pointer to an array and <code>p[x][y]</code> is an item in the pointed-to array.</p>
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