Little vs Big Endianess: How to interpret the test

Question

So I\'m writing a program to test the endianess of a machine and print it. I understand the difference between little and big endian, however, from what I\'ve found online,

Answer 1

int x;

x is a variable which can hold a 32-bit value.

int x = 1;

A given hardware can store the value 1 as a 32-bit value in one of the following format.

Little Endian
0x100    0x101    0x102    0x103
00000001 00000000 00000000 00000000 

(or) 

Big Endian
0x100    0x101    0x102    0x103
00000000 00000000 00000000 00000001

Now lets try to break the expression:

&x 

Get the address of variable x. Say the address of x is 0x100.

(char *)&x 

&x is an address of an integer variable. (char *)&x converts the address 0x100 from (int *) to (char *).

*(char *)&x 

de-references the value stored in the (char *) which is nothing but the first byte (from left to right) in the 4-byte (32-bit integer x).

(*(char *)&x == 1)

If the first byte from left to right stores the value 00000001, then it is little endian. If the 4th byte from left to right stores value 00000001, then it is big endian.