Assignment of an integer variable address to character pointer

Question

Following is a piece of C++ code which assigns integer variable address to character pointer.

int ia = (1<<25) + (1<<24) + (1<<17) + (1<<8) + 4;
char *cArr = (char *)(&ia);
int i;
for(i = 0; i < 4; i++){ cout << (int)cArr[i] << endl; }

I have chosen the integer variable so that the MSB byte can have a value of 3, the next value of 2, the next a value of 1 and the LSB byte a value of 4.

Now when I assign the integer's address to a character pointer then I get the following result:

cArr[0] = 4
cArr[1] = 1
cArr[2] = 2
cArr[3] = 3

I have some questions regarding this result:

1. Is it always that the LSB byte will get the first address irrespective of the machine on which I am working.
   
2. Has the result something to do with Little Endian and Big Endian. Also I am looking for some basic tutorial which can get me the basics of Little Endian and Big Endian.
   
3. Has the result something to do with how the stack grows on a machine.
   
   

By the way I work on a windows PC

Answer 1

1). No that depends upon the computer architecture and hardware. Here is the table that can be taken as a reference.

2). Basic tutorial. Hope this would suffice. http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Data/endian.html

3).No nothing to do with stack as you see from table windows follow, Little Endian and hence LSB place first.

let me know if that answers you.

Bye!

Answer 2

1. No. The results depends on your system hardware type. Only little-endian machines get your results.

2. Just as question 1, this indeed is determined by little-endian or big-endian. Here is a comprehensive introduction on endian. http://www.yolinux.com/TUTORIALS/Endian-Byte-Order.html

3. There is nothing to do with stack in your case.