CODEWITHSUNDEEP
cpointersmultidimensional-array
Amol Singh
Amol Singh

Reputation: 161

Please explain the difference

i have a program about 2-D arrays

base adress is 8678

#include<stdio.h>
#include<conio.h>
main()
{
 int arr[3][3]={
                 {83,8,43},
                 {73,45,6},
                 {34,67,9}
                 };
printf("%d ",&arr+1);  //points to 8696
printf("%d ",arr+1);   //points to 8684
return 0;
}

what is the difference between arr+1 and &arr+1?

Upvotes: 5

Views: 994

Answers (4)

Gdbvhv Cxb
Gdbvhv Cxb

Reputation: 1

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

Upvotes: 0

hupantingxue
hupantingxue

Reputation: 2234

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

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

Upvotes: -1

Jim Balter
Jim Balter

Reputation: 16406

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.

Upvotes: 1

Carl Norum
Carl Norum

Reputation: 224864

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.

Upvotes: 9

Related Questions