Boolean Algebra - A little help please...

MobyDick · 3 Mar 2011 at 12:39

ok so we're given an example of " 0xAA && 0x12 = 0x01 ". This I worked out and got the same, all good.

Next example is " !0xAA && 0x12 = 0x00 ". Now I worked this out to be 0x01 :confused:

Code:

0xAA = 10101010;
!0xAA = 01010101;

0x12 = 00010010;

so    01010101
      00010010 &&
      00010000

which would be 0x01 :confused:

Am I going wrong somewhere or is the example wrong?

MobyDick · 3 Mar 2011 at 13:07

This is the way it's shown:

MobyDick · 3 Mar 2011 at 13:11

DaveF said:
(I don't see how you could have got 0xAA & 0x12 = 0x01" under the bitwise interpretation at all).

I got

Code:

0xAA = 10101010
0x12 = 00010010
       00000010

so 0x01

MobyDick · 3 Mar 2011 at 15:13

DaveF said:
OK, then those are 'C' style operators. You don't work in a bitwise fashion.

As far as your bitwise calculation of 0xAA & 0x12: your final answer is 2 (00000010), not 1 (00000001).

It says the && are logical operators. "For logical operators any non zero value is considered 1." That's why I got 1, not 2.

MobyDick · 3 Mar 2011 at 15:42

Ok just found something I had written in the note. "Logical. Don't convert to binary". So do it evilpauls way? Must remember to look through my notes

MobyDick · 3 Mar 2011 at 16:48

DaveF said:
But in that case you should have considered 0xAA and 0x12 as 1 as well.

Hence the distinction between bitwise operators and logical operators.

I realise that now

Thanks for helping me out everyone!