Cast void* to struct*

Question

I have this struct:

typedef struct
{
    UINT8 a;
    UINT8 len;
    BYTE *data;
} MyStruct;

and this binary array [0x00, 0x03, 0x08, 0x09, 0x0a] which is assigned to a void* variable "BINDATA".

How can I cast BINDATA to MyStruct and be able to access its "data" field?

I tried:

MyStruct *myStruct = (MyStruct*) BINDATA;

After that I was able to access:

myStruct->a; //gave me 0x00
myStruct->len; //gave me 0x03

But I could not access

myStruct->data;

without memory access violation. I guess this is because "data" address pointer gets set to 0x08 and not its value.

What is ```myStruct->b```? in your struct I cant see b member. — , Aug 12 '16 at 13:39
Undefined behavior for accessing an object through an lvalue not of the effective type of the object. — EOF, Aug 12 '16 at 13:40
What is `myStruct->data` supposed to be? What is the correct result for this example? — ForceBru, Aug 12 '16 at 13:41
`myStruct->data` shall be a pointer to BYTE with following contents: `0x08, 0x09, 0x0a` — arminb, Aug 12 '16 at 13:42
It's doable with some changes to MyStruct and some packing directives, but the way it is now, you would have to make a new MyStruct and set the pointer to the correct address for "data", as well as initialize the other values. — Kenny Ostrom, Aug 12 '16 at 13:43
Are you sure pointers are 24 bit on the given system? 16, 32 or 64 bit pointers are much more common. — Lundin, Aug 12 '16 at 13:46
Show code saving a struct to BINDATA and loading from it. It must be error there. — Anton Malyshev, Aug 12 '16 at 13:50
This violates [strict aliasing](http://stackoverflow.com/questions/98650/what-is-the-strict-aliasing-rule), and so EOF is right in that it's undefined behavior. — AndyG, Aug 12 '16 at 13:50

score 4 · Accepted Answer · answered Aug 12 '16 at 13:42

4

Instead of BYTE *data you should use BYTE data[0] or (if your compiler doesn't like this) BYTE data[1]. The difference between pointer and array here is crucial - array is "data that is right here" while pointer is "data somewhere else" which is not your case.

answered Aug 12 '16 at 13:42

aragaer

17,238
6
47
49

Wow this actually works. But isn't that a little hacky? It looks weird in the code to define data[1]; – arminb Aug 12 '16 at 13:48
1

@arminb, not at all. You're just putting data in an array, which is OK. – ForceBru Aug 12 '16 at 13:49
2

@arminb It's called a "flexible array member". It's allowed in C, but not in C++. – Barmar Aug 12 '16 at 13:54
@Barmar how do you mean it's not allowed? How do you do this in C++? – arminb Aug 12 '16 at 13:56
@arminb In C, you declare the flexible array member with an empty size `BYTE data[]`. That's what isn't allowed. If you declare it with a size, nothing stops you from allocating more space with `malloc()` and writing more elements. But the language doesn't address the proper way to calculate the additional size. – Barmar Aug 12 '16 at 13:59
Also C++ `new MyStruct` can't be used to allocate larger space. – Barmar Aug 12 '16 at 14:00
In C++ you can use `(MyStruct *) ::operator new(sizeof(MyStruct)+len)` – aragaer Aug 12 '16 at 14:17

Captain Dirty Pants · Answer 2 · 2016-08-12T14:52:22.263

-1

Something like this may work in c++:

UINT* a1 = &myStruct->len;
UINT* a2 = ++a1;

Then cast a2 to whatever you want. Be aware of the size and the type of your data.

edited Aug 12 '16 at 14:52

answered Aug 12 '16 at 14:38

Captain Dirty Pants

1
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Cast void* to struct*

2 Answers2