C Programming

Question

#include<stdio.h>
int main()
{
    int a[2][2][2]={10,2,3,4,5,6,7,8};
    int *p,*q;
    p=&a[1][1][1];
    q=(int*)a;  //Please explain this line
    printf("%d%d"*p,*q);
    return 0;
}

Why we are doing typecasting here and what will be returned after type cast to q?

Comments

'a' is initially equivalent to triple dimensional pointer i.e ***a and p is single dimensional pointer i.e *p

now if you want to perform p=a, you have to cast 'a' into single dimensional pointer, that is what (int*)a do.

q=(int*)a;  will simply make q to point to first element of array i.e '10', it is equivalent to q=&a[0];

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in this line 

q=(int*)a;

q is storing the starting address of a 

hence *q will print 10

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@joshi_nitish is it correct to say that,when normally when we do typecast like p=(int)a; it wil return value a,now here since it is given like p=(int *)a; this will return a pointer to pointer to a i.e currently p is refring to a[0][0].