output

Question

#include<stdio.h>

int main() {
int array[2][3]={5,10,15,20,25,30};
int (*ptr)[2][3]=&array;
printf("%d\t",**(*ptr+1));
}
can someone explain how is it "20"??

Comments

*(ptr+0) will point to first row 

*(ptr+1) will point to second row

**(*ptr+1) will give the value at that location i.e. second row first column.

Answer 1

let array 

a[2][3]={5,10,15,20,25,30};  here a will hold the address of initial elements of array....

here *a=a  and **a=*(*(a+1)+0)=a[1][0] will print 20....means 1^th row and 0^th column....

now in this case...

 (*ptr)[2][3]=&a ....then  *ptr=a and then like above.....
 

,**(*ptr+1) also can be written as *(*(*ptr+1)+0) or *(*(a+1)+0) or a[1][0].....

Answer 2

In computer a 2D array is stored in row major order and ptr is a pointer which points to the row of array means ptr+0 points to first row...ptr+1 points to second row so *ptr+1 points to the address of  second row....and**(*ptr+1) will give the value stored at*ptr+1 i.e.,20

Answer 3

Hope this diagram helps you to understand.