Can someone explain what's happening in printf expression??

Question

https://code.sololearn.com/ck69IP47w8p8/?ref=app

Answer 1

(Part 2)

Lets imagine that
variable = *(arr+1);
variable is now equivalent to the {4,5,6} array (in this example)

And accessing the numerical elements will be no different than a regular array.

To get the third element
*(variable+2) == 6

Now replace variable with the original code:
*(*(arr+1)+2)
This is getting the third element of the second array.

and in your code the i and the j are the iterating offsets in a loop

Answer 2

(Part 1)
In arrays, using addresses and pointers, elements are accessed by deferencing.

e.g.
int arr[3] = {1,2,3};
printf("%d",*(arr+1));
//This will print the second element, 2. The addend is the address offset of the array. +0 = first element, +1 = second element etc...

Remember that a multidimensional array is an array of arrays, i.e. an array that holds arrays.
Array: {1,2,3}
Array of arrays: {{1,2,3}, {4,5,6}, {7,8,9}};

These individual arrays are accessed the same way as a regular array.

So look at this piece by piece
*(*(arr+i)+j)

*(arr+i)
This piece of code is referring to the elements in the multidimensional array, where the elements are the single dimensional arrays.

When deferencing this, the element we recieve is a single dimensional array
*(arr+1) = {4,5,6}
So this returns a single dimensional array.

So now we've establish that *(arr+1) in this situation is returning a single dimensional array.

Answer 3

Thanks for clarification.Nice explanation.

Answer 4

Martin Taylor According my professor
"A fool can't think like a computer and solve the code" 😅😅

Answer 5

Yeah yeah I know this.But the code was an academical question.As it was the given question I have to sove it.

Answer 6

Martin Taylor