Algorithm for bitwise fiddling

Question

If I have a 32-bit binary number and I want to replace the lower 16-bit of the binary number with a 16-bit number that I have and keep the upper 16-bit of that number to produce a new binary number.. how can I do this using simple bitwise operator?

For example the 32-bit binary number is:

1010 0000 1011 1111 0100 1000 1010 1001

and the lower 16-bit I have is:

                    0000 0000 0000 0001

so the result is:

1010 0000 1011 1111 0000 0000 0000 0001

how can I do this?

Answer 1

You do this in two steps:

• Mask out the bits that you want to replace (AND it with 0s)
• Fill in the replacements (OR it with the new bits)

So in your case,

i32 number;
i32 mask_lower_16 = FFFF0000;
i16 newValue;

number = (number AND mask_lower_16) OR newValue;

In actual programming language implementation, you may also need to address the issue of sign extension on the 16-bit value. In Java, for example, you have to mask the upper 16 bits of the short like this:

    short v = (short) 0xF00D;
    int number = 0x12345678;
    number = (number & 0xFFFF0000) | (v & 0x0000FFFF);
    System.out.println(Integer.toHexString(number)); // "1234f00d"

Answer 2

(original32BitNumber & 0xFFFF0000) | 16bitNumber

Answer 3

Well, I could tell you the answer. But perhaps this is homework. So I won't.

Consider that you have a few options:

| // bitwise OR
^ // bitwise XOR
& // bitwise AND

Maybe draw up a little table and decide which one will give you the right result (when you operate on the right section of your larger binary number).

Answer 4

use & to mask off the low bits and then | to merge the 16 bit value with the 32 bit value

 uint  a = 0xa0bf68a9
 short b = 1

 uint  result = (a & 0xFFFF0000) | b;