My program is good enough for my assignment but I know its not good

Question

I'm just starting an assignment for uni and it's raised a question for me.

I don't understand how to return a string from a function without having a memory leak.

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;
}

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:

char trimmed[len];

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.

So instead I made my array by allocating some memory to a char pointer.

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:

1. Can you normally initialise an array using a variable to declare the length like:

   char trimmed[len];

2. 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*).

3. 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.

Answer 1

To address (3) - You can free the newly allocated string from your calling code, once you are finished with it:

char* tmp = trim(myline);

if (tmp != NULL) {
    ....
    free( tmp );
}

But this places a burden on the caller to remember to free the memory. So instead you might consider passing an allocated buffer and a buffer size to trim(), for example:

void trim(char* line, char *trimmed_buf, int trimmed_buf_len){ ... }

Neil did an excellent job addressing your other questions. Basically an array declaration char trimmed[len]; will declare a local variable on the stack, so while it is syntactically correct to return a char * to this memory, the memory location that it points to will no longer be valid.