Struct name ptr = (Same struct name ) p2;

Question

What is the meaning of this statement in C?

struct foo *ptr = (struct foo *) p2;

***"Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking."*** — abelenky, Jan 28 '14 at 13:30
Well it is clearly what he is asking, but it is just unanswerable without additional code. — Jori, Jan 28 '14 at 13:31
Without seeing some information about the type and value of p2, we cannot answer. Your comment: *if (ptr && p2->variable)* makes no sense in relation to the code you posted. — abelenky, Jan 28 '14 at 13:32

score 2 · Answer 1 · answered Jan 28 '14 at 13:33

2

You're not giving all the informations we need:

struct name *ptr = (Same struct name *) p2;

let's make it something that can compile:

struct foo* ptr = (struct foo*) p2;

but now we're missing what p2 is. So, I'm going to assume that p2 is a plain C pointer, i.e. void*:

void* p2;
struct foo* ptr = (struct foo*) p2;

So here you're assigning to ptr the address pointed by p2. In my example now, this is pretty pointless… but if you allocate some memory:

void* p2 = malloc(sizeof(struct foo));
struct foo* ptr = (struct foo*) p2;

then p2 is having the address of a memory space that then you assign to ptr.

The example I'm giving you here is using two variables for doing the usual:

struct foo* ptr = (struct foo*) malloc(sizeof(struct foo));

answered Jan 28 '14 at 13:33

zmo

UINT32 p2; is the type – Fagun Jan 28 '14 at 13:36
1

Since void* can be assigned to any pointer casting examples are kinda irrelevant. – this Jan 28 '14 at 13:37
[looks like it is considered to be](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) though I've been taught to cast my mallocs at uni, and I think there are some compilers complaining about this when you do `-Wall`. – zmo Jan 28 '14 at 13:37
No C compiler should ever complain about casting from `void*` – David Heffernan Jan 28 '14 at 13:38
Not if they are standard compliant. Casting is only useful when using C++ compiler for C code, which is a bad idea to begin with. – Jori Jan 28 '14 at 13:38
ok, good to know, Jori. Though there are some cases where you have a C++ compiler and the only way to allocate memory is to use `malloc()` (as new is not existing). I'm thinking of AVR embedded development. But I'd agree that's an exception… – zmo Jan 28 '14 at 13:38

David Heffernan · Accepted Answer · 2014-01-28T13:47:18.940

2

Starting at the left hand side:

struct foo *ptr

declares ptr to be of type struct foo *, that is a pointer to struct foo.

The = initializes that ptr variable to whatever the right hand side evaluates as.

And the right hand side

(struct foo *) p2

is an expression that casts p2 to be of type struct foo *.

edited Jan 28 '14 at 13:47

answered Jan 28 '14 at 13:35

David Heffernan

Struct name *ptr = (Same struct name *) p2;