How Could I Store A Struct In Char A Byte Array

Question

I have a byte array created like this in my C source file:

static char arr[64];

I also have a struct declared like so:

static char arr[64];

struct test {
    int foo;
    int data;
};

If everything in memory is just bytes, how could I store the bytes of of a test struct inside of arr

I have tried the following things:

int main() {
    struct test t;
    t.foo = 255;
    t.data = 364;

    arr[0] = t; // This did not work; I got a type-mismatch

    // I also tried memcpy

    memcpy(arr, &t, 8); // But this did not work either because it does not store the data in array. I was also not able to deference the bytes that did get stored.
}

Is there any easy way that I can store the bytes of the test struct in the byte array arr so that I can store multiple structures in this array and access it easily too? If everything is just bytes, is there a feasible way to store the test struct bytes in the arr array?

You're getting into potential undefined behavior. See [**What is the strict aliasing rule?**](https://stackoverflow.com/questions/98650/what-is-the-strict-aliasing-rule) — Andrew Henle, Feb 15 '22 at 13:46
@AndrewHenle I thought using `memcpy` was safe against aliasing and alignment requirements. — Adrian Mole, Feb 15 '22 at 13:49
@Darth-CodeX The name of an array used in an expression or as a function parameter is automatically converted (decays) into a pointer to its first element. — Adrian Mole, Feb 15 '22 at 13:50
@Darth-CodeX `arr` is declared globally (outside of main). So, it would be stored in global space. — NewRProgrammer, Feb 15 '22 at 13:53
`memcpy(arr, &t, 8);` or rather `memcpy(arr, &t, sizeof(struct test));` itself is fine, but what are you doing with `arr` later? — Jabberwocky, Feb 15 '22 at 13:57
Using `sizeof(struct test)` instead of the `8` constant should work. Maybe an `int` is bigger than 4 bytes on your platform? — Adrian Mole, Feb 15 '22 at 13:57

AKX · Answer 1 · 2022-02-15T13:54:54.450

1

You can:

#include <stdio.h>
#include <string.h>

struct test {
    int foo;
    int data;
};
int main() {
    struct test t;
    char arr[sizeof(struct test)];
    t.foo = 0x12345678;
    t.data = 1364;
    memcpy(arr, &t, sizeof(struct test));
    printf("%02x %02x %02x %02x\n", arr[0], arr[1], arr[2], arr[3]);
    printf("%02x %02x %02x %02x\n", arr[4], arr[5], arr[6], arr[7]);
}

This prints out

78 56 34 12
54 05 00 00

(or thereabouts, depending on endianness and struct alignment).

edited Feb 15 '22 at 13:54

answered Feb 15 '22 at 13:49

AKX

152,115
15
115
172

Note that `struct test *t = (struct test *)arr;` **is** a strict-aliasing violation. – Andrew Henle Feb 15 '22 at 13:51
@AndrewHenle Revised with 100% more memcpy. – AKX Feb 15 '22 at 13:55

score 0 · Answer 2 · answered Feb 15 '22 at 14:35

It would seem that you are simply looking for this:
(assuming no struct padding ever present)

struct test {
  int foo;
  int data;
};

static struct test array [8] =
{
  { .foo = 123, .data = 456 }
  ...
};

Then for any random chunk of data of any type, C allows us to inspect it with a character pointer:

unsigned char* raw_data = (unsigned char*)&array;
...
raw_data[i] // access individual bytes

So the separate 64 byte array fills no purpose.

Alternatively, you could do this:

typedef struct {
  int32_t x;
  int32_t y;
} foo_t;

typedef union {
  foo_t   foo   [8];
  uint8_t bytes [64];
} bar_t;

Again, this is assuming no padding (a struct with just 2 int won't have any).

How Could I Store A Struct In Char A Byte Array

2 Answers2