How to declare and initialize an array of random bytes in C?

Question

I can create an array of bytes like below:

test_array[3] = {0x01, 0x02, 0x03};

I want an array of bytes like above, but with random bytes. For example, I just declare that I need an array of 10 bytes. and then I get the array of size 10, but all random bytes. So that at the end, when I write:

printf("My array is: %d", test_array);

I need to see this:

{0x09, 0x15, 0xA1, 0xB2, 0xF1, 0x33, 0xBC, 0xCA, 0x1B, 0x9D};

You need to fill the array yourself with random values. 5 seconds of googling revealed e.g this: https://stackoverflow.com/questions/822323/how-to-generate-a-random-int-in-c — Jabberwocky, Oct 20 '22 at 08:39
BTW: `printf("My array is: %d", test_array);` won't work. You cannot print an entire array, you need to do it yourself using a loop. C is not Python. I suggest you get some good beginner's C text book. — Jabberwocky, Oct 20 '22 at 08:41
I need to fill more than 1000 bytes in my array, then filling it manually will be difficult so I was looking that maybe there exist a function so that it will be filled with random bytes. But as you mentioned it doesn't seem to be possible. Can we also fill it using loop maybe? — Raza Javed, Oct 20 '22 at 09:10
You obviously need a loop here. This is most basic general progamming knowledge. — Jabberwocky, Oct 20 '22 at 09:20

score 0 · Answer 1 · answered Oct 20 '22 at 09:19

I need to fill more than 1000 bytes in my array, then filling it manually will be difficult so I was looking that maybe there exist a function so that it will be filled with random bytes.

5 lines of code ...

unsigned char *create(void *data, const size_t size)
{
    unsigned char *ucdata = data ? data : malloc(size * sizeof(*ucdata));

    if(ucdata)
    {
        for(size_t i = 0; i < size; i++)
        {
            ucdata[i] = rand();
        }
    }
    return ucdata;
}

usage:

int main(void)
{
    unsigned char arr[5000];
    unsigned char *x;
    
    srand(time(NULL));
    create(arr, sizeof(arr));
    x = create(NULL, 150000); 

    /* ... */
    free(x);
}

So that at the end, when I write:

printf("My array is: %d", test_array);

I need to see this: {0x09, 0x15, 0xA1, 0xB2, 0xF1, 0x33, 0xBC, 0xCA, 0x1B, 0x9D};

This printf for sure will not display it. You need to iterate, but you need to do something yourself

How to declare and initialize an array of random bytes in C?

1 Answers1