What does assigning a value to a function that returns a pointer do in C? *function=value

Question

So I am looking at this and the guy is continuously using

uint32_t *internalNodeNumKeys(void *node)
{
    return (uint32_t *)(node + INTERNAL_NODE_NUM_KEYS_OFFSET);
}

*internalNodeNumKeys(root) = 1;

What does this do? Somewhere I remember him saying that because these functions return pointers they can be used as setters but what do they set?

That's not a function pointer; that's a function that returns a pointer. — Govind Parmar, Feb 03 '19 at 22:49
`node + INTERNAL_NODE_NUM_KEYS_OFFSET` is [not legal](https://stackoverflow.com/questions/3523145/pointer-arithmetic-for-void-pointer-in-c). — jamesdlin, Feb 03 '19 at 22:56

chqrlie · Accepted Answer · 2019-02-03T23:15:41.433

2

The function returns a pointer to int, the expression *internalNodeNumKeys(root) = 1; is parsed as:

*(internalNodeNumKeys(root)) = 1;

Postfix operators such as () for function call bind more tightly than prefix operators such as * for dereferencing.

Note also that internalNodeNumKeys is highly non portable:

Performing pointer arithmetics on void pointers is non portable, node should be cast as (unsigned char *) before the addition.
Casting an arbitrary offset into the object pointed to by node as a pointer to int may have undefined behavior, the programmer is playing with fire at this point.

edited Feb 03 '19 at 23:15

answered Feb 03 '19 at 22:57

chqrlie

Yes, but as it has been addressed in one of the comments to the question by Jamesdlin, `node + INTERNAL_NODE_NUM_KEYS_OFFSET` is illegal, as it is doing pointer arithmetic with a `void *`. It should be better to have declared the parameter as `char *` instead. – Luis Colorado Feb 05 '19 at 04:53

1 Answers1