Pointer to different data type in C

Question

I have compiled and run the following program in C:

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>

int main(int argc, char* argv[]){
  uint8_t data[8] = {0, 1, 2, 3, 4, 5, 6, 7};
  uint32_t* pointer32 = &data[0];
  uint64_t* pointer64 = &data[0];

  printf("%" PRIu64 "\n", *pointer64);
  printf("%" PRIu32 "\n", *(pointer32++));
  printf("%" PRIu32 "\n", *pointer32);

  return 0;
}

and received the following expected output:

506097522914230528
50462976
117835012

The output is correct and corresponds to the bitwise interpretation of the data as an unsigned 64-bit integer and unsigned 32-bit integers. I tried this on a 64-bit machine running Ubuntu 14.04. It was compiled with the stock gcc compiler (4.8.4?). The compiler does throw the "assignment from incompatible pointer type" warning (which can be safely ignored because the incompatible assignment is intended).

Is this a reliable way of converting and interpreting the data in the "data" array, or would it be better recommended to manually copy and shift each byte, one at a time, to a temporary variable?

"which can be safely ignored"... – Kerrek SB Jul 08 '15 at 19:30 — Kerrek SB, Jul 08 '15 at 19:30
It will have different behaviour for different endianness. – Eugene Sh. Jul 08 '15 at 19:31 — Eugene Sh., Jul 08 '15 at 19:31

too honest for this site · Accepted Answer · 2015-07-08T19:42:54.270

5

You are violating aliasing rules. So, the clear answer is: no.

Briefly: you must not have pointers of different type pointing to the same object. This may result in broken code, as the compiler actually does assume this does not happen and might optimize the code.

Just a very strong hint: do not ignore warnings. They are given for good reasons.

The best way is to properly serialize/deserialize the data from the array per element to the final types. This will also avoid any problems with endianess (byte-ordering) of the values and any (possible) padding.

I use functions like this:

uint32_t readUInt32(const uint8_t **buffer)
{
    ... // increment buffer accordingly
}

This way, I just pass the buffer pointer along the line, without having to care about incrementing in the caller. The technic is actually an iterator.

edited Jul 08 '15 at 19:42

answered Jul 08 '15 at 19:31

too honest for this site

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@haccks: Just leave me a second;-) – too honest for this site Jul 08 '15 at 19:34
Hahahaaaaaaaa. Take your time. – haccks Jul 08 '15 at 19:34
-1, since this seems unaware that `char*` is a special case, and that might be relevant here. (and if it's not relevant, that deserves explanation) – Jul 08 '15 at 20:00
1

@Hurkyl: No, I did not oversee that! You OTOH might have not noticed OP assigns the `uint8_t *` to **two** pointers of different types! **This** creates the aliasing failure, not the assignment of the `uint8_t *`. Read the link I provided and you will see that. – too honest for this site Jul 08 '15 at 20:06

score 1 · Answer 2 · answered Jul 08 '15 at 19:34

The compiler does throw the "assignment from incompatible pointer type" warning (which can be safely ignored because the incompatible assignment is intended).

This warning cannot and should not be ignored, because the pointers are indeed incompatible. The reason this is problematic is that interpretation of data depends on the hardware in use - specifically, on the endianness of the machine.

would it be better recommended to manually copy and shift each byte, one at a time, to a temporary variable?

This approach would be independent of the endianness. Alternatively, you could use hton{...}/ ntoh{...} family of functions to force the byte order.

score 0 · Answer 3 · answered Jul 08 '15 at 19:35

0

The behavior will differ between a big-endian machine and a little-endian machine, so you're better off manually copying and shifting each byte.

answered Jul 08 '15 at 19:35

dbush

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Pointer to different data type in C

3 Answers3