Please explain the difference

Question

i have a program about 2-D arrays

base adress is 8678

#include
#include
         


        

Answer 1

Arr+1 gives the next element in an array while &arr +1 gives the address of next array of integers

Answer 2

Well, they're different things. arr decays in most contexts to a pointer to the first element of your array - that means a pointer to the first 3-element row in your 2D array: type int (*)[3]. arr + 1, then, points to the second row in the array.

&arr is the address of the array itself (type int (*)[3][3]), so &arr + 1 points to memory just past the end of the entirety of your 2D array.

You can confirm this behaviour easily by printing differently. Specifically, printing the offsets to the new pointers rather than the values themselves will help clear things up. The output from your program from these print statements:

printf("%ld\n",(intptr_t)(&arr+1) - (intptr_t)arr);
printf("%ld\n",(intptr_t)(arr+1) - (intptr_t)arr);

Will be the decimal offsets to &arr+1 and arr+1 respectively. Here's the output from a test run I just made:

36
12

36 matches up: 3 rows × 3 columns × 4 bytes per entry = 36 bytes. So does the 12: 1 row × 3 columns × 4 bytes per entry = 12 bytes.

Note - you're also printing pointers using %d, which is wrong. You should probably be using %p for that.

Answer 3

You can figure this out with the help of this equivalence: X[Y] === *(X+Y)

Since *(arr+1) === arr[1], arr+1 === &arr[1]

Similarly, &arr+1 === &((&arr)[1])

What is (&arr)[1]? Well, (&arr)[0] === *&arr === arr, that is, the 3x3 array itself, so (&arr)[1] is the 3x3 array following that, and &arr+1 === &((&arr)[1]) is the address of the 3x3 array following &arr ... a pointer to the byte just past the entire array.

Answer 4

array + 1 means the array[1] 's address and it costs 3 int memory.

&array + 1 means the address of array[0] add 1;