How to follow C++ bit storage in Java?

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...

You might be talking about endian-ness. Is endian-ness not a property of the system? Can you really change endian-ness? — Sriram, May 09 '11 at 11:36
@Sriram, as @unapersson says, this doesn't have anything to do with Endianess, given the data structure in C++ is an *array* of `char`s rather than say an `int` — Nim, May 09 '11 at 11:38
btw - `abc[32] = 0xC0204420` is wrong, if you want to initialize - you can do something like (C++) `unsigned char abc[] = {0xC0, 0x20, 0x44, 0x20};`. — Nim, May 09 '11 at 11:44

score 1 · Answer 1 · answered May 09 '11 at 11:34

1

Yes, that will be how C++ represents the char array in memory.

As for the Java, it's completely arbitrary. It depends on how you define the mapping between the index into your Bitset and the byte/bit index in the C++ "representation". You can define this mapping any way you like, so long as you're consistent.

answered May 09 '11 at 11:34

Oliver Charlesworth

267,707
33
569
680

1

AFAIK, there is no direct way to set a group of bits in the `BitSet`, so this is a manual, bit-then-byte iteration operation... – Nim May 09 '11 at 11:42

score 1 · Accepted Answer · answered May 09 '11 at 11:42

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.

score 0 · Answer 3 · answered May 09 '11 at 11:45

If you create char abc[4]; and fill it character by character, then yes, in memory it will be represented as 11000000 00100000 01000100 00100000.

The problem I see with Java BitSet is that the contents of a BitSet cannot be represented simple as a number - there is no method for it, so how would you convert 0xC0204420 to a BitSet or how would you represent a BitSet as a number like 0xC0204420? One possibility is to take every eight bits from a BitSet, make a char from them (using Java bitwise operators like << and |) and then create some char array or something from them that should be equal to the earlier C++ char[4].

How to follow C++ bit storage in Java?

3 Answers3