Answer of below code is 5. How _**(x+i+j)_ indicates x[3][0] memory location contain. Explain.

Question

#include <stdio.h>

int main (){

int x [4][5],i,j,k;

for (i=0; i<4 ; ++i)
for (j=0,k=1; j<5 ; j++){

x[i][j] = k;

k++;
}

i=1;
j=2;

x[3][0] = 5;

printf("%d",**(x+i+j));

return 0;

}

Answer 1

the code creates a two dimensional array 4 x 5:
12345 first row
12345 2nd row
12345 3rd row
12345 4th row
think of it like a table.
x[3][0] means 4th row first column. (array index starts from 0.)
x[3][0] = 5;
now the array look like :
12345 first row
12345 2nd row
12345 3rd row
52345 4th row (notice 5)
array name can be used as a pointer to the first element when it's dereferenced.
incrementing x inside **( ) will go through rows.
ie: **(x + 1) will point to the first value in 2nd row.
so **(x + 1 + 2) will point to the first element in 4th row.
because it's changed by x[3][0] = 5
it equals 5.

if it were **(x) + i + j; i and j outside of **()
it would mean 1 + 1 + 2