Question

if I have in c++:

char abc[4];

abc[0] = 0xC0; //11000000 in binary

abc[1] = 0x20; //00100000 in binary

abc[2] = 0x44; //01000100 in binary

abc[3] = 0x20; //00100000 in binary

So how this will be stored in memory –

11000000 00100000 01000100 00100000 or the reverse way ??
----------------------------------- 
   0th      1st     2nd      3rd

In Java I am creating Bitset abc = new Bitset(32);

So I need to store the same values in this(same order).This may be
modified later according to bit positions so have to be exact same way.

So abc[32] = 0xC0204420 will do? And if I want to store
the values in c++ way what to do??
If I am wrong then how to do this in Java…

Answer 1

Endian is not an issue. If you use char[4] the lowest address 0 will be first, the highest 3 will be last, so you get in memory

char[0] char[1] char[2] char[3]

whatever you do.

If you do a int x = *(reinterpret_cast<int*>(abc)), then you get different results, depending on endianness, because the (4byte-) int is sometimes read as 0123, sometimes 3210 — and I think even 2301 has been around in the 60s.

You can not put 0xC0204420 (a larger number then 127) into the [32]ths position of abc. If you want to implement something “fast” (and dangerous) you would need a platform-depending reinterpret_cast. Take a look at hton and ntoh.