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copied!<p>There is no need to empty it. Often in C, memory allocation is done with <code>malloc</code> which simply returns to you a block of memory which is deemed owned by the caller. When <code>calloc</code> is called, as well as returning you a block of memory, the memory is guaranteed to be initialized to 0. This means for all intents and purposes it is already 'empty'.</p> <p>Also I'm not quite sure if your code does what you are intending. Let me explain what it does at the moment:</p> <pre><code>char arg_array = (char**)calloc(strlen(buf), sizeof (char**)); </code></pre> <p>This line is simply wrong. In C, there is no need to cast pointers returned from <code>calloc</code> because they are of type <code>void *</code>, which is implicitly casted to any other pointer type. In this case, you are storing it in a <code>char</code> type which makes no sense. If you do this:</p> <pre><code>char ** arg_array = calloc(strlen(buf), sizeof (char**)); </code></pre> <p>Then it allocates an array of pointers of <code>strlen(buf)</code> length. So if <code>buf</code> is <code>"hello"</code> then you have now allocated an array which can store 5 pointers.</p> <pre><code>for(i = 0; i<(strlen(buf)); i++) { arg_array[i] = calloc (strlen(buf), sizeof(char*)); } </code></pre> <p>Again, I have removed the redundant cast. What this does is populates the array allocated earlier. Each index of the array now points to a <code>char</code> string of <code>strlen(buf) * sizeof(char *)</code> length. <strong>This is probably not what you want.</strong></p> <hr> <p>Your question is more clear to me now. It appears you want to remove the strings after populating them. You can do it two ways:</p> <ul> <li>Either <code>free</code> each of the pointers and allocate more space later as you did before</li> <li>Or set the first character of each of the strings to a null character</li> </ul> <p>To free the pointers:</p> <pre><code>for(i = 0; i<(strlen(buf)); i++) { free(arg_array[i]); } </code></pre> <p>To set the first character of each string to a null character:</p> <pre><code>for(i = 0; i<(strlen(buf)); i++) { arg_array[i][0] = '\0'; } </code></pre> <p>That is the same code as what you have originally and should be fine.</p> <hr> <p>As proof, the following code will run without errors:</p> <pre><code>#include <stdlib.h> #include <stdio.h> #include <string.h> int main(void) { char * buf = "hello"; char ** arg_array = calloc(strlen(buf), sizeof (char**)); unsigned int i; for(i = 0; i < strlen(buf); i++) { arg_array[i] = calloc(strlen(buf), sizeof(char *)); } for(i = 0; i < strlen(buf); i++) { arg_array[i][0] = '\0'; } for(i = 0; i < strlen(buf); i++) { free(arg_array[i]); } free(arg_array); return EXIT_SUCCESS; } </code></pre> <p>If your code is segfaulting, the problem is coming from somewhere else. Did you overwrite the <code>arg_array</code> variable? Are you sure <code>BUFSIZE</code> is equal to <code>strlen(buf)</code>?</p>

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