C: pointer to array != pointer to &array[0]?

Question

Suppose I have two arrays:

uint8_t buf1[10];
uint8_t buf2[20];

I now want to create an array of these arrays, eg,

uint8_t *buffers[] = {buf1, buf2};

When I look at the buffers variable in the debugger, however, I see that its two elements do not point to any elements of buf1 and buf2 , but look like random values (maybe the contents of buf1 / buf2 ?). Also, when dereferencing, eg,

uint8_t x = *(buffers[0] + 1);

the programs throws a bus error.

If I change the definition to

uint8_t *buffers[] = {&buf1[0], &buf2[0]};

everything is alright.

Why is there a difference? I always thought that array and &array[0] are equivalent?! What is the correct type of buffers to make definition 1 work?

EDIT: OK, I goofed up. There is in fact no difference between definition 1 and 2. I'm not really sure what change I did to see the correct values, but I do have a screenshot.

Anyway, I now know the answer to my question why the pointers are all bad, but you couldn't know because I stripped an important detail from the question.

I added __attribute__((section(".foo"))) to the definition of buffers , and that segment foo is uninitialized . Of course the values are random!

Sorry for wasting your time. :-)

Answer 1

At onlinegdb the following code:

#include <stdint.h>

uint8_t buf1[10];
uint8_t buf2[20];
uint8_t* buffers1[] = {buf1, buf2};
uint8_t* buffers2[] = {&buf1[0], &buf2[0]};

int main()
{
    uint8_t x1 = *(buffers1[0] + 1);
    uint8_t x2 = *(buffers2[0] + 1);

    return 0;
}

at the return statements has the following state:

Which seems to do what you expect, but not what you describe. How does this test differ from yours?