array long with pointers

Question

give the complete solution with explanation

int main()
{
    int arr [2] [2] [2] = {10,2,3,4,5,6,7,8};
    int *p, *q;
    p = &arr[1] [1] [1];
    q = (int*) arr;
    printf("%d ,%d \n",*p ,*q);
    return 0;

}

Comments

8 and 10

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@shivam sharma 5

don't miss use the tags.... you can include your requirement either in your question or comment

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please provide the stepwise solution so that it can be easy to understand

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a[1][1][1] so first a[1] land you to the matrix {5,6,7,8} now you need to choose a[1][1] of this matrix so it will be 8 and  and  when we assign the array name to a pointer we give the base address to it so *q will give 10 hence result will be 8 10

Answer 1

Ans will be

*p = 8

*q = 10

for clear idea, your array can be seen as

arr[0]            arr[1]

$\begin{bmatrix} 10 & 2\\ 3& 4 \end{bmatrix}$   $\begin{bmatrix} 5 & 6\\ 7& 8 \end{bmatrix}$

Comments

 p = &arr[1] [1] [1]

p holds the address of the element at 1^th column of 1^th row of 1^th 2-D array.

1^th 2-D matrix is $\begin{pmatrix} 5 &6 \\ 7 &8 \end{pmatrix}$

Then 1^th row is [7 8]

Element at 1th colum is 8.

So p holds the address of  8 and *p gives value at that address which is 8.

q holds the base address of the 3-D array type casted to (int *) as q is a pointer to an integer type. The element at the base is 10(i.e. the first element). So *q gives 10.