Error in multidimensional arrays in cpp

help
1

Why does the following code snippet fails at runtime?
It should work because the memory addresses are contiguous( i even tried by printing memory addresses)

void print(int **mat, int r, int c){
   for(int i=0;i<r;++i){
        for(int j=0;j<c;++j){
            cout << mat[i][j] << " ";
        }
        cout << endl;
    }
}

int main(){
    int mat[2][2] = { {1,2}, {3,4} };
    print( (int **)mat,2,2);
}
2
1 Like
3

I got itā€¦but my question is "since i am collecting the base address of the 2d array in the print function, and then i am accessing the elements using pointer arithmetic and it should work since the elements are consecutiveā€¦but why doesnā€™t it work?? "

4

When you write
int mat[2][2] = {{1, 2}, {3, 4}};
Note that mat[0] is a constant pointer, pointing to the address of the element in the first row and first column (i.e., 1).
Similarly, mat[1] is a constant pointer, pointing to the address of the element in the second row and first column (i.e., 3).

Now, when you write
int ** pointers = (int **) mat,
the variable pointers is a ā€œpointer to a pointerā€. What happens this way is that the address of the first element in the first row will be assigned to the variable pointers. And remaining addresses are not assigned.

You can check it using the following code:

#include <bits/stdc++.h>
using namespace std;

void print(int **mat, int r, int c){
   cout << mat;
}

int main(){
    int mat[2][2] = { {1,2}, {3,4} };
    cout << &mat[0][0] << '\n';
    cout << mat << '\n';
    print( (int **)mat,2,2);
}

// All three cout statements print the same hex value

Now that other addresses are not known to the variable pointers, it will result in an \color{red}\text{RTE}.

1 Like
5

/*to be honest i tried the same problem and wasted hoursšŸ˜‘. i guess the programm will work if converted in c languageā€¦but here is an alternative

#include
using namespace std;

void print(int mat, int r, int c){
for(int i=0;i<rc;++i){
cout << *(mat+i) << " ";
if(i+1==r)
cout<<endl;
};
}

int main(){
int mat[2][2] = { 1,2,
3,4 };
/* int **p;
p=(int **)mat;
cout<<p<<endl;
cout<<++p<<endl;*/
print(mat[0],2,2);
}

6

Correct me if i am wrong,

let say the base address of matrix is 2000, then
&mat[0][0] = 2000
&mat[0][1] = 2004
&mat[1][0] = 2008
&mat[1][1] = 2012

Now when we access mat[i][j] in main method, it is equivalent to saying * ( *(mat+i) + j) , where
mat+i points to ith array in matrix as type of mat is int ( * )[2]. (mat+1 will be pointing to address 2008)

Now when we try to access mat[i][j] in print function, *( *(mat+i) + j) will not yield mat[i][j] since the base type of mat is int ** , mat+i will no longer point to ith array.
(In this case, mat+1 will be pointing to 2004)

This is right, isnā€™t it?