How is each byte in an integer stored in CPU / memory?

Question

i have tried this

char c[4];
int i=89;
memcpy(&c[0],&i,4);
cout<<(int)c[0]<<endl;
cout<<(int)c[1]<<endl;
cout<<(int)c[2]<<endl;
cout<<(int)c[3]<<endl;

the output is like:
89
0
0
0

which pretty trains my stomache cuz i thought the number would be saved in memory like 0x00000059 so how come c[0] is 89 ? i thought it is supposed to be in c[3]...

Answer 1

Because the processor you are running on is little-endian. The byte order, of a multi-byte fundamental type, is swapped. On a big-endian machine it would be as you expect.

Answer 2

This is because you're running the program on a little endian cpu. See also endianness here and there.

Answer 3

Endian-ness is obviously the answer as Goz has pointed out.

But for those who are unclear of what that means, it's also important to understand that the order of bytes displayed in the example is the same as the order in the original int. Memcpy doesn't change the byte order, regardless of the edian type of the platform.