My char pointer points to invalid value after being cast from int*

Question

I am learning C programming language, I have just started learning arrays with pointers. I have problem in this question, I hope the that output must be 5 but it is 2, Can anyone please explain why?

int main(){    int arr[] = {1, 2, 3, 4, 5};    char *ptr = (char *) arr;    printf("%d", *(ptr+4));    return 0; } 

Answer 1

Assumed a little endian architecture where an int is 32 bits (4 bytes), the individual bytes of int arr[] look like this (least significant byte at the lower address. All values in hex):

|01 00 00 00|02 00 00 00|03 00 00 00|04 00 00 00|05 00 00 00 

char *ptr = (char *) arr; 

Now, ptr points to the first byte - since you have casted to char*, it is treated as char array onwards:

|1|0|0|0|2|0|0|0|3|0|0|0|4|0|0|0|5|0|0|0  ^  +-- ptr 

Then, *(ptr+4) accesses the fifth element of the char array and returns the corresponding char value:

|1|0|0|0|2|0|0|0|3|0|0|0|4|0|0|0|5|0|0|0          ^          +-- *(ptr + 4) = 2 

Hence, printf() prints 2.

On a Big Endian system, the order of the bytes within each int is reversed, resulting in

|0|0|0|1|0|0|0|2|0|0|0|3|0|0|0|4|0|0|0|5          ^          +-- *(ptr + 4) = 0