c - Why do we need to use double pointer to access a 2-D array? -

i trying understand multidimensional array , pointers, wrote small program understand concept:

#include<stdio.h> void show(int arr[][2]); int main() {     int z[2][2] = { { 1, 2 },                     {3, 4 } };     show(z); }  void show(int arr[][2]) {      printf("value of arr = %d", arr);     printf("\n\nvalue of &arr[0]0] = %d", &arr[0][0]); } 

this code fragment prints same address makes sense, when edit show function:

void show(int arr[][2]) {      printf("value of *arr = %d", *arr);     printf("\n\nvalue of arr[0]0] = %d", arr[0][0]); } 

*arr still prints same address while arr[0][0] expected prints integer value , want know why need use **arr int value , if arr storing address should dereferenced *arr, isn't ?

please having hard time understanding concept.. in advance.

if @ memory layout of 2d array, things might become little bit clearer.

you have variable defined as:

int z[2][2] = {{1, 2}, {3, 4}}; 

memory:

z | v +-----+-----+-----+-----+ |  1  |  2  |  3  |  4  | +-----+-----+-----+-----+ 

another view of memory:

z[0]        z[1] |           | v           v +-----+-----+-----+-----+ |  1  |  2  |  3  |  4  | +-----+-----+-----+-----+ 

another view of memory:

z[0][0]     z[1][0] |   z[0][1] |    z[1][1] |     |     |     | v     v     v     v +-----+-----+-----+-----+ |  1  |  2  |  3  |  4  | +-----+-----+-----+-----+ 

you can see that, far pure memory location concerned,

&z == &z[0] == &z[0][0] 

we know when array decays pointer, value address of first element of array. hence, when used in expression decays z pointer,

z == &z[0] == &z (from above) 

it's puzzling z , &z evaluate same address though of different types.

type of z when decays pointer int (*)[2]. type of &z int (*)[2][2].

coming function, have:

void show(int arr[][2]) { ... } 

that equivalent to:

void show(int (*arr)[2]) { ... } 

why arr , *arr evaluate same value?

arr evaluates &z[0] main. *arr evaluates z[0] main, evaluates &z[0][0] main.

we have seen value of &z[0] , &z[0][0] same. hence arr , *arr in show() evaluate same address.