CODEWITHSUNDEEP
c++arrayspointers
shivam mitra
shivam mitra

Reputation: 232

Address of an element in 2-d array

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)?

Upvotes: 0

Views: 164

Answers (2)

Akash Mahapatra
Akash Mahapatra

Reputation: 3248

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].

Upvotes: 2

Sumeet
Sumeet

Reputation: 8292

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.

Upvotes: 1

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