On Wed, 23 Feb 2005, wrote:

> So how do I take the complement of an unsigned number that has 1 as
> the MSB? If I'm using an 8 bit system, and a = 10000101b = 133d, what
> is a's complement? Is it 01111010?

There are two terms: "two's complement representation" and
applying "two's complement" operation (~a+1)=-a without worrying about
the semantics of whether it is signed or unsigned.

10000101b = a

01111010b = ~a = one's complement
+      1b
---------
01111011b = (~a+1) = two's complement

===============================================

Some more questions about homework #4, second page:
1) The statement u = a - b. If a and b are both unsigned (they have to
be if we're using 8 bits, since their values are between 127 and 255)
and b is greater than a, how do I perform a subtraction here? The
result is negative, but I'm working with unsigned numbers.


In two's complement: u = a - b is the same as u = a + (~b + 1)

Don't need subtract instruction.


2) Same question for s = w - x.

The same as above. You will notice in the 8051 machine instructions
that there are no signed add or unsigned add only add. The 8051
just add bits and does not care about semantics as a C compiler does.

3) Similarly, in s = ~w and s= -w, how do I get the complement if w is
an unsigned number and the MSB is 1?

~w is always 1's complement. Again, once you have the bits just "not"
them. And -w is ~w+1


4) For the last one, s = -z ^ ~a, I got a as the answer. Is this
correct? Also, how would I code this in 8051?

mov a,r6  ;regA=z;
cpl a     ;regA = ~regA
add a,#1  ;regA = regA + 1
mov r3,a  ;s=regA

mov a,r0  ;regA=a
cpl a     ;regA=~regA
xrl a,r3  ;regA=regA ^ s
mov r3,a  ;s=regA

===============================================

> Could you explain the difference between big endian and little endian?
> I understand that the one's complemement of 01100010 is 10011101 and

yes

> the two's complement is 10011111 (right?),

no,
      10011101
    +        1
      --------
      10011110

where 1+1 is Sum=0 and Carry=1

see page 32, table 2-3

 but where do big endian and
> little endian come into play with this?

big "bit" endian is most significant bit first:

dn-1 .... d3 d2 d1 d0 = 128+64+...+4+2+1 = 10011110

little "bit" endian is least significant bit first:

d0 d1 d2 d3 .... dn-1 = 1+2+4+...+64+128 = 01111001

=============================================

--Prof. Wolff.