Address of an element in 2-d array

Question

for(i = 0; i < n; i++) {
    for(j = 0; j < n; j++) {
        printf("%u ",(*(a+i)+j));
    }
    cout<<endl;
}

Assuming that 2-d array a is declared, how (*(a+i)+j) gives the address of each element? I am not understanding this concept. Usually *(a+i) should give the value at location (a+i)?

Akash Mahapatra · Accepted Answer · 2016-11-12 17:28:45Z

2

Here it is the case of Row Major ordering where the 2-D array is stored row wise.
Assuming the indexing starts from 0, to access any element in the array, say A[i][j], we have to first cross "i" number of rows and then "j" number of columns.

Since 'a' is the base address of the array, *( a + i ) gives the location in the memory after i rows ,i.e ith row and adding j to it takes the pointer to the jth column of the ith row giving the address of A[i][j].

answered Nov 12, 2016 at 17:28

Akash Mahapatra

3,3281 gold badge16 silver badges28 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Sumeet · Accepted Answer · 2015-06-26 16:09:46Z

1

yes *(a+i) gives the value at that location which is pointer to the beginning of the ith row and + j takes the pointer from the beginning of the row to the jth element of the ith row.

So i is taking care of the rows and j is taking care of the columns.

Hence (*(a+i) + j) is giving the address of each element. Hope this clears.

answered Jun 26, 2015 at 16:09

Sumeet

8,3745 gold badges27 silver badges51 bronze badges

Collectives™ on Stack Overflow

Address of an element in 2-d array

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related