Challenge: Index an element of an n-dimensional array using pointers arithmetic.

Question

You can not use moltiplication!!!

For example:

int matrix[20][20];
matrix[10][5] = 30;
//Now print matrix[10][5] using pointers arithmetic
cout << *((int*)matrix+(10*20)+5) << endl;
cout << *((int*)matrix+(200)+5) << endl;
//Both are not valid for this challenge (10*20 is a moltiplication)

//Then try with a 3-dimensional array, always without moltiplications

int t[10][15][20];
t[8][12][16] = 30;
//Now print t[8][12][16] using pointers arithmetic.

Answer 1

moltiplication is a repeated addition...so no you cant...you need another tipe of addition ;)

Answer 2

Can we use repeated addition? 😅
I don't think so...

Answer 3

No i am sorry. If you create a matrix using pointers-to-pointers you are right. But try with a normal matrix (int matrix[20][20]) and a normal 3d array (int t[10][15][20])

Answer 4

XD
now in 3d

Answer 5

Sure:

#include<iostream>
using namespace std;
int main()
{
int mat[9][9][9];
mat[4][7][1]=30;
mat[1][2][3]=33;
cout<<*(*(*(mat+1)+2)+3);
//Format - *(*(*(mat+slice)+row)+column)
}

Answer 6

Good.

Answer 7

Last question:
mat is not a *** but a (*)[9][9] so why triple dereferentiation work? 😂😂😂
How the compiler consider it? What does it really do?

Answer 8

The operator[] is overloaded in the following way, thats why...

T being a custom type...

T operator[](int index)
{
return *(arr+index);
}

Now, C++ has defined these operators and created arrays etc from pointers. So they are simply the same thing.

But int*** a is != int*[9][9], as the second one has a fixed number of elements, but the *** one is resizable...

Answer 9

I win!

#include<iostream>
using namespace std;
int main()
{
int** mat=new int*[9];
for(int i=0;i<9;i++) mat[i]=new int[9];
mat[4][7]=30;
mat[7][4]=33;
cout<<*(*(mat+7)+4);
//Format - *(*(mat+row)+column);
}

Answer 10

@Andrea
Is this ok? :)

#include<iostream>
using namespace std;
int main()
{
int mat[9][9];
mat[4][7]=30;
mat[7][4]=33;
cout<<*(*(mat+4)+7);
}