Array Matrix

[lozovic]

int matrix[2][2], i, j;
for (i=0; i < 2; i++) {
int *ptr = matrix[0] + 2*i;
for (j=0; j < 2; j++)
ptr[j] = (i+1)*(j+1);
}

hey guys, i got across this in one of my past paper questions, however I do not seem to be able to figure this out. Have tried searching for array in matrix but dont seem to understand.

I am suppose to state the value of the array matrix after the code is executed.

thx!!!

[hamsterman]

If you don't know what that value is, compile it. Is there something you don't understand about this code?

[lozovic]

i tried, but i keep getting all sorts of errors, i do not get the matrix, *ptr and ptr bit
thx

[hamsterman]

#include <iostream>

int main(){
    int matrix[2][2], i, j;
    for (i=0; i < 2; i++) {
        int *ptr = matrix[0] + 2*i;
        for (j=0; j < 2; j++)
            ptr[j] = (i+1)*(j+1);
    }
    for(i = 0; i < 2; i++) {
        for(j = 0; j < 2; j++)
            std::cout << matrix[i][j] << " ";
        std::cout << '\n';
    }
    std::cin.ignore();
    return 0;
}

Added output.

Are you familiar with pointers at all?
matrix is a 2x2 array. ptr is a pointer. The thing to understand here is that even though you call it a 2d array, memory is always linear. Every element of the array can be accessed using simple pointer arithmetics.

[lozovic]

no I was not, but i've just read quite a bit abt it now. thx a lot!!
now that you've added the the output part, i am completely confused, also for what is the differnce between ptr and *ptr, as i could not find much about this particular pointer function.

Again thx a lot, as i am only a beginner in C++

[hamsterman]

line 6 (in my code) sets ptr to ith row of matrix.
Here * is only there to show that ptr is a pointer. You could do

typedef int* int_pointer;
...
int_pointer ptr;

It's a part of the type.

In other cases, * returns the value a pointer points to.

int a = 5;
int* ptr = &a;
cout << ptr << '\n';//prints 0xSOMETHING
cout << *ptr;//prints 5;