Implicit pointer cast in parameter and argument of functions

Question

If 2 function definitions are:

    void func(struct Node *arg){...}

    void func2(void *arg){
      func(arg);
      ...
    }

but they are called like:

    struct Node *node = (char *)malloc(6);
    func2(node)

I think node is implictly casted to void* and then to struct Node*, so I don't need to do something like:

    func2((void *)node);
    or func((struct Node *)arg);

Is my understanding correct?

Not related, but why `struct Node *node = (char *)malloc(6);`, instead of `struct Node *node = (struct None *) malloc(6);`? — informatik01, May 22 '13 at 16:46
Or [better](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc), `struct Node *node = malloc(6);` — interjay, May 22 '13 at 16:47
@H2CO3 Yup, was just asking out of curiosity. btw nice thread here: [Do I cast the result of malloc?](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) — informatik01, May 22 '13 at 16:49
@informatik01 If you check my answer, that link is included :) — , May 22 '13 at 16:50

score 3 · Accepted Answer · edited May 23 '17 at 12:28

There's no such thing as "implicit casting". There is explicit type conversion (casting) and implicit type conversion (type coercion or promotion).
Since void * is compatible with any data pointer type (and if your implementation conforms to POSIX, then it's compatible with function pointers too), your assumption is right:

T *object = malloc(size);

is right regardless to the type T denotes. The same applies to function arguments, of course.

1 Answers1