allocating node on the heap, difference between first assign to NULL and direct assignment with mall-CodePudding

I am trying to implement a stack using linked list, and firstly I have the following code:

typedef struct Node {
    int data;               // integer data
    struct Node* next;      // pointer to the next node
} Node;


Node* inti_stack() {
    Node* node = NULL;// allocate a new node in a heap
    node = malloc(sizeof * node);
    if (!node) exit(EXIT_FAILURE);
    return node;
}

For the inti_stack function, can I just do the following and that would be equivalent?

Node* inti_stack() {
    Node* node = malloc(sizeof * node);
    if (!node) exit(EXIT_FAILURE);
    return node;
}

CodePudding user response：

In the first case:

Node* node = NULL;// allocate a new node in a heap
node = malloc(sizeof * node);

The value of node is determinate. malloc() will overwrite it with a pointer to a chunk of heap memory if it succeeded.

In the second case:

Node* node = malloc(sizeof * node);

The value of node is indeterminate, but it's going to be overwritten by malloc() nevertheless.

So yes, they're both equivalent. Though the latter is preferred.

NB that malloc() will return a NULL pointer on failure, and set errno on some implementations. Your code should check for it.

Aside: Why exit() without an error message?

if (!node) exit(EXIT_FAILURE);
/* I'd expect a fprintf()/perror() here */

CodePudding user response：

In the first code snippet

Node* node = NULL;// allocate a new node in a heap
node = malloc(sizeof * node);

the declared pointer node is at once overwritten.

So it is enough to write

Node* node = malloc(sizeof * node);

If the function malloc is unable to allocate memory it returns a null pointer.

Pay attention to that the name of the function inti_stack is unclear.