Re: signed unsiged add instruction

Gianmaria wrote:

There is no difference in signed/unsigned addition if you
use 2's complement representation for signed numbers.

that's clear.

If that's clear, then I don't understand your question below.


Suppose (just for simplicity) to work with 3 bit long word (000, 001
etc. range -4..3) and you want to add two signed number: 100 (-4) and
111 (-1). The result -5 is to big to be encoded in the 3 bit signed
word.
I was thinking that in case you use signed add instruction this will
preserve the sign bit (the leftmost bit) and give the result 111. On
the contrary if you use a unsigned add I was thinking the result would
be 011. So, if I understand correctly what I'm saying here is
perfectly incorrect :) Isn't it?

You want to add the binary numbers 111 + 011. (NOTE: add/sub is
always done modulo 2^n, in this case modulo 8)

The ALU doesn't know whether this bit sequences are signed
or unsigned numbers so it has to add them without this information:

100
+111
-----
011

It's up to the user to interpret this binary numbers. If you
want them to be unsigned numbers this means: 4 + 7 = 3
Or for signed numbers this means: -4 + -1 = 3
Both interpretations are correct (don't forget, we do modulo 8
arithmetic).

Processors which uses flags, signals an overflow (wrapping around
modulo 8) by setting one of the flags. Because the CPU doesn't know
whether the binary number should represent a signed or unsigned
number, it has to use two different flags to signal an overflow.
The C flag signals on overflow if the binary numbers represent
an unsigned number and the V flag if the number is a signed number.
So in the example above C=1 and V=1.
.

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