Split 64-bit value to four 16-bit

Question

I have 64 bit values like this:

1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 0011

Decimal value of given stream of number is -45.

I want to split in four 16-bit values:

1111 1111 1111 1111
1111 1111 1111 1111
1111 1111 1111 1111
1111 1111 1101 0011

The input value is signed long.

How to do that in Java?

So far I've tried on this way:

long[] buf = new long[4];
buf[0] =  (l&0xFF);
buf[1] =  (l&0xFF >> 16) & 0XFF;
buf[2] =  (l&0xFF >> 32) & 0XFF;
buf[3] =  (l&0xFF >> 48) & 0XFF;

But I think that I mess something with bit shifting logic.

EDIT

The correct solution thanks to @Eugene is:

long l = 45;

long[] buf = new long[4];
buf[0] =  (l & 0xFFFF);
buf[1] =  l  >> 16 & 0XFFFF;
buf[2] =  l  >> 32 & 0XFFFF;
buf[3] =  l  >> 48 & 0XFFFF;

long result = ( buf[3] & 0xFFFF) << 48 | ( buf[2] & 0xFFFF) << 32 | (buf[1] & 0xFFFF) << 16 | (buf[0] & 0xFFFF);

System.out.println("Decimal: " + result);
System.out.println("Binary: "Long.toBinaryString(result));

Answer 1

EDIT

you are almost correct (I did way too much typing thinking that you needs the basics also, my bad).

All you have to change is: buf[1] = l >> 16 & 0xFFFF;

I could have written it differently, but that is probably the easiest to understand:

 long t = Long.parseUnsignedLong("1111111111111111111111111111111111111111111111111111111111010011", 2);

    long twoPower16 = Long.parseLong("1111111111111111", 2);

    long one = t & twoPower16;
    long two = t >> 16 & twoPower16;
    long thr = t >> 32 & twoPower16;
    long fou = t >> 48 & twoPower16;

    System.out.println(Long.toBinaryString(one));
    System.out.println(Long.toBinaryString(two));
    System.out.println(Long.toBinaryString(thr));
    System.out.println(Long.toBinaryString(fou));

When you do an & operation(it's called a mask), for example :

   0101
   0011
   ----
   0001 --> you are keeping the last two bits here 

and the >> is a shift operator. You tell how many bits to shift:

 System.out.println(7 >> 1); // 3

That's actually 0111 >> 1 --> 0011 , you have shifted 7 on bit to the left.