How to detect if memory was allocated? (Linked lists)

Question

I want to create a function that adds an element to the end of a linked list. It also has to return 0 if the element was added successfully, or a 1 if memory could not allocated/saved to for the element.

The question is, how do I know if the memory was allocated successfully or if the element was added successfully? This is the code:

int push_back(pos_t *head, int new_value) {
    pos_t *temp = head;

    while (temp->next != NULL) {
        temp = temp->next;
    }
    pos_t *temp1 = (pos_t *)malloc(sizeof(pos_t));
    temp1->data = new_value;
    temp1->next = NULL;
    temp = temp1;
}

Is head == NULL possible? It would be useful to see how the link-list is initialized. — chux
– chux, Commented Oct 23, 2017 at 21:49

Edwin Buck · Accepted Answer · 2017-10-23 21:35:22Z

1

You need to add the following code

if (temp1 == NULL) { return 1; }

because malloc is defined to either return a pointer to the allocated memory, or

if the size is zero, return NULL.
on error, return NULL.

You can control that you don't request a size of zero, so if you used a positive size, and malloc returned NULL, you can deduce an error occurred.

Many systems have "manuals" installed. If you are using a Linux system, the command "man malloc" will pull up the manual page for malloc. If you are working on a Windows system, a web search for the manual for malloc will give you enough detail to handle the details.

answered Oct 23, 2017 at 21:35

Edwin Buck

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1 Comment

Edwin Buck Over a year ago

Of course, you'll need to add a return 0; too, and the locations of both code additions are left for you to figure out.

chqrlie · Accepted Answer · 2017-10-23 21:50:02Z

The function has a shortcoming: it cannot be used to allocate the first node in the list, ie if the list is empty.

The prototype should be changed to

int push_back(pos_t **headp, int new_value);

passing the address of the list pointer instead of its value.

Testing for malloc() failure is simple: juts compare the returned pointer to NULL or 0.

Here is the corresponding code:

int push_back(pos_t **headp, int new_value) {
    pos_t *temp = *headp;
    pos_t *temp1 = malloc(sizeof(pos_t));
    if (temp1 == NULL) {   // allocation failure
        return 1;
    }
    temp1->data = new_value;
    temp1->next = NULL;

    if (temp == NULL) {      // empty list
        *headp = temp1;
    } else {
        while (temp->next != NULL) {
            temp = temp->next;
        }
        temp->next = temp1;  // append node to the end of the list
    }
    return 0;
}

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