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POMy program is good enough for my assignment but I know its not good
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copied!<p>I'm just starting an assignment for uni and it's raised a question for me.</p> <p>I don't understand how to return a string from a function without having a memory leak.</p> <pre><code>char* trim(char* line) { int start = 0; int end = strlen(line) - 1; /* find the start position of the string */ while(isspace(line[start]) != 0) { start++; } //printf("start is %d\n", start); /* find the position end of the string */ while(isspace(line[end]) != 0) { end--; } //printf("end is %d\n", end); /* calculate string length and add 1 for the sentinel */ int len = end - start + 2; /* initialise char array to len and read in characters */ int i; char* trimmed = calloc(sizeof(char), len); for(i = 0; i < (len - 1); i++) { trimmed[i] = line[start + i]; } trimmed[len - 1] = '\0'; return trimmed; } </code></pre> <p>as you can see I am returning a pointer to char which is an array. I found that if I tried to make the 'trimmed' array by something like:</p> <pre><code>char trimmed[len]; </code></pre> <p>then the compiler would throw up a message saying that a constant was expected on this line. I assume this meant that for some reason you can't use variables as the array length when initialising an array, although something tells me that can't be right.</p> <p>So instead I made my array by allocating some memory to a char pointer.</p> <p>I understand that this function is probably waaaaay sub-optimal for what it is trying to do, but what I really want to know is:</p> <ol> <li><p>Can you normally initialise an array using a variable to declare the length like:</p> <p>char trimmed[len];</p></li> <li><p>If I had an array that was of that type (char trimmed[]) would it have the same return type as a pointer to char (ie char*).</p></li> <li><p>If I make my array by allocating some memory and allocating it to a char pointer, how do I free this memory. It seems to me that once I have returned this array, I can't access it to free it as it is a local variable.</p></li> </ol>

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