How does the function return something without return statement?

Question

It works in the Windows Command Prompt like I haven't missed the return new_node;

#include <stdio.h>
#include <stdlib.h>

typedef struct Node {
    int value;
    struct Node *next;
 } Node;

Node* create_node(int value) {
    Node *new_node = (Node *) malloc(sizeof(Node));
    if (new_node == NULL) return NULL;
    new_node->value = value;
    new_node->next = NULL;
    // no return statement
}

int main(int argc, char *argv[]) {
    Node *head = NULL;

    // no errors here, head just receives the right pointer
    head = create_node(5);

    return 0;
}

So the function create_node(int) returns pointer anyway. How does it work?

Compiled with gcc (x86_64-posix-seh-rev1, Built by MinGW-W64 project) 7.2.0

It's undefined behavior. It just looks like it is returning something. Accidentally. — Eugene Sh., Mar 02 '18 at 18:03
That is because you did not enable warnings of your compiler: it would have warned you. — Paul Ogilvie, Mar 02 '18 at 18:03

user2736738 · Accepted Answer · 2018-03-02T18:05:25.463

That's undefined behavior and standard mentions it clearly

From §6.9.1¶12 C11 standard

If the } that terminates a function is reached, and the value of the function call is used by the caller, the behavior is undefined.

Compile your code with all warnings enabled. gcc -Wall -Werror prog.c. In this case you will see compiler mentioning that there is no return statement though it is supposed to return something.

How does the function return something without return statement?

1 Answers1