Set digit of a hexadecimal number

Question

How can I set a digit in a hexadecimal number? I currently have this code:

int row = 0x00000000;
row |= 0x3 << 8;
row |= 0x2 << 4;
row |= 0x1 << 0;

printf("Row: 0x%08x", row);

Which works perfectly fine as long as "row" is just zeros. As soon as I change it to something like this:

int row = 0x33333333;
row |= 0x3 << 8;
row |= 0x2 << 4;
row |= 0x1 << 0;

printf("Row: 0x%08x", row);

I just get this output:

Row: 0x33333333

Answer 1

You should delete (make it 0) the digit first.

row &= ~(0xf << 4);

~ operator reverses the values of all bits in the number. So. 0x000000f0 becomes 0xffffff0f.

Your code should look like:

row &= ~(0xf << 8);
row |= 0x3 << 8;
row &= ~(0xf << 4);
row |= 0x2 << 4;
row &= ~(0xf << 0);
row |= 0x1 << 0;

Answer 2

As Alexandru explained, you do need to clear the bitfield you're trying to set before you go on to set it.

Just to add further comment on why your code didn't do what you wanted, consider what is happening to the variable at the binary level:

row |= 0x2 << 4;

The |= operator is a "bitwise or". Hence if either the bit you're trying to set OR the bit you're passing in is set to 1, the result is 1. In your code, row is set to 0x33333333, so each 4 bit hexadecimal digit is 0011 in binary. When you bitwise or that with 0x2, you get 0x3:

/* 0x3 | 0x2 = 0x3 */
0011 | 0010 = 0011

If you clear the bitfield first, you get 0x2, which is what you want:

/* 0x3 | 0x0 = 0x0 */
0011 | 0000 = 0000

/* 0x0 | 0x2 = 0x2 */
0000 | 0010 = 0010

Note that data manipulation using shifts and bitwise operations is unlikely to be portable between different platforms. You may run into problems trying to run code relying on bit shifts on machines of different endianess, or indeed if you try to run code that works on a 32 bit machine on a 64 bit machine:

http://www.viva64.com/en/a/0004/#ID0EQFDI

Wikipedia has more on bitwise operations:

http://en.wikipedia.org/wiki/Bitwise_operation