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2011-08-19T19:23:01.700
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The thing that confuses me about your question is that most people would not declare a matrix as an <code>int **</code>. The reason for this is that you would be forced to then allocate it in a loop. Your allocation function would require two parameters, which are the dimensions of the array like this: <pre><code>void AllocMat(int *** mat, int n, int m) { int ** result = new int * [ n ]; for (int x=0; x<n; x++) { result[x] = new int [ m ]; } *mat = result; } </code></pre> If this were the case, the corresponding deallocation function would require knowledge of the size of n as follows: <pre><code>void DeallocMat(int *** mat, int n) { if (mat == NULL || *mat == NULL) return; int ** tmp = *mat; for (int x=0; x<n; x++) { if (tmp[x] != NULL) delete [] tmp[x]; } delete [] tmp; *mat = NULL; } </code></pre> With this approach, you could access your matrix like this: <pre><code>int ** mat = NULL; AllocMat(&mat, n, m); for (int x=0; x<n; x++) { for (int y=0; y<m; y++) { mat[x][y] = 1; } } DeallocMat(&mat, n); </code></pre> Usually, people allocate matrices as a single buffer of memory to avoid extra allocations and pointer indirections, which is how I recommend you do it. In that case, you allocation function would look like this: <pre><code>void AllocMat2(int ** mat, int n, int m) { *mat = new int [ n * m ]; } </code></pre> And the corresponding deallocation function like this: <pre><code>void DeallocMat2(int ** mat) { if (mat != NULL && *mat != NULL) { delete [] *mat; *mat = NULL; } } </code></pre> And you would access it follows: <pre><code>int * mat2 = NULL; AllocMat2(&mat2, n, m); for (int x=0; x<n; x++) { for (int y=0; y<m; y++) { mat2[x * n + y] = 1; } } DeallocMat2(&mat2); </code></pre>
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