CODEWITHSUNDEEP
javabit-manipulationbitwise-operators
Aviram
Aviram

Reputation: 3067

Reverse bytes order of long

I've got a long variable and I need to reverse its byte order. For example: B1, B2, ... , B8 I should return a long that consists of B8, B7, ..., B1. How can I do it by using bitwise operations?

Upvotes: 4

Views: 8090

Answers (5)

sheldonzy
sheldonzy

Reputation: 5961

Plain answer with loops:

public static long byteReverse(long a) {

    long result = 0;
    for(int i = 0; i < 8; i++){
        // grab the byte in the ith place
        long x = (a >> (i*8)) & (0b11111111);
        result <<= 8;
        result |= x;
    }
    return result;
}

bitwise only:

public static long byteReverse(long a) {

    a = (a << 32) | (a >>> 32);
    a = ((a & 0xffff0000ffff0000L) >>> 16) | ((a & 0x0000ffff0000ffffL) << 16);
    a = ((a & 0x00ff00ff00ff00ffL) << 8) | ((a & 0xff00ff00ff00ff00L) >>> 8);

    return a;
}

Upvotes: 0

user1964389
user1964389

Reputation: 1

long reverse(long x){
     x = (x >> 32) | (x << 32); // step 1
     x = ((x & 0xffff0000ffff0000 ) >> 16) 
          | ((x & 0x0000ffff0000ffff ) << 16); // step 2
     x = ((x & 0xff00ff00ff00ff00 ) >> 8) 
          | ((x & 0x00ff00ff00ff00ff ) << 8); // step 3
     return x;
}

If we assume that bitwise operator works in O(1) time, reverse function works in O(lg(number of bits) ) time.

Explanation
Step 0 : B1 B2 B3 B4 B5 B6 B7 B8
Step 1 : B5 B6 B7 B8 B1 B2 B3 B4
Step 2 : B7 B8 B5 B6 B3 B4 B1 B2
Step 3 : B8 B7 B6 B5 B4 B3 B2 B1

Upvotes: 0

Serdalis
Serdalis

Reputation: 10489

you can use Long.reverseBytes(long)

Or for more methods which include bitwise operations, you can refer to this stack overflow question

Heres another method you may like, I'd still recommend the above but it's better than bitwise where you can easily make mistakes.

Bytebuffer

byte[] bytes = ByteBuffer.allocate(8).putLong(someLong).array();
for (int left = 0, right = bytes.length - 1; left < right; ++left, --right) {
    byte temp = bytes[left]; 
    bytes[left]  = bytes[right]; 
    bytes[right] = temp;
}

I am trying to steer you away from bitwise solutions because they are cumbersome and very easy to mess up if you do not know what you are doing... But bitwise would look like this:

byte[] bytes = new byte[8];

// set the byte array from smallest to largest byte
for(int i = 0; i < 8; ++i) {
    byte[i] = (your_long >> i*8) & 0xFF;
}

// build the new long from largest to smallest byte (reversed)
long l = ((buf[0] & 0xFFL) << 56) |
         ((buf[1] & 0xFFL) << 48) |
         ((buf[2] & 0xFFL) << 40) |
         ((buf[3] & 0xFFL) << 32) |
         ((buf[4] & 0xFFL) << 24) |
         ((buf[5] & 0xFFL) << 16) |
         ((buf[6] & 0xFFL) <<  8) |
         ((buf[7] & 0xFFL) <<  0) ;

Upvotes: 6

vhallac
vhallac

Reputation: 13927

Here's an old trick that you can use to endian swap a register:

static long swapblock(long a, long mask, int shift) {
    long b1 = a & mask; // extract block
    long b2 = a ^ b1;   // extract remaining bits
    return (b1 << shift) |
           ((b2 >> shift) & mask); // mask again to clear sign extension
}

static long endianswap(long a) {
    a = swapblock(a, 0x00000000ffffffffL, 32);
    a = swapblock(a, 0x0000ffff0000ffffL, 16);
    a = swapblock(a, 0x00ff00ff00ff00ffL, 8);
    return a;
}

The idea is to progressively swap sub blocks until you reach the desired level you want to stop at. By adding swaps of sizes 4, 2, and 1, you can change this into a bit mirror function.

There is only one tricky bit due to lack of unsigned types in java. You need to mask out high order bits when shifting right, because the sign bit is replicated by the shift amount, filling the high order bits with ones (0x8000000000000000 >> 8 is 0xFF80000000000000).

Upvotes: 1

Matthias
Matthias

Reputation: 7521

You might want to use Long.reverseBytes instead of using bitwise operations. See the Java Reference for details.

Otherwise, you could have a look at the JDK sources (src.zip in your JDK folder) in Long.java but mind the copyright by Oracle.

Upvotes: 2

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