In 16 bit 2s complement signed arithmetic:

0111 1111 1111 1111  =3D +32767
0111 1111 1111 1110  =3D +32766
..
..
..
0000 0000 0000 0001  =3D +1
0000 0000 0000 0000  =3D 0
1111 1111 1111 1111  =3D -1
1111 1111 1111 1110  =3D -2
..
..
..
1000 0000 0000 0001  =3D -32767
1000 0000 0000 0000  =3D -32768

So, adding 1 to +32767 results in -32768.  No number can be larger than=20
+32767 or smaller than -32768.
To change the sign of a number, e.g. -10 to +10, invert all the bits and=20
add 1.

Does that help?

Kerry






IVP wrote:
> Hi all, and hi to my brain, wherever it went for Christmas. Come
> back soon, miss you
>
> I'm trying to rationalise this testing and limiting to express it in=20
> assembler
>
> If Result > 32767 then Result =3D 32767
>
> If Result < -32768 then Result =3D -32768
>
> It would help if I could get an idea of how more-negative works
> past 16 bits. I've not found anything specifically useful after some
> looking around
>
> In this example, -32768 - (+20)
>
>            mov     #20,w0              ;temp0 =3D 20 =3D 0x0014
>            mov     w0,temp0
>            mov     -#32768,w1          ;temp1 =3D -32768 =3D 0x8000
>            mov     w1,temp1
>
>            mov     temp0,w0
>            mov     temp1,w1
>            sub     w0,w1,w2            ;w2 =3D 0x8014 =3D -32748
>
> In this example, -32768 - (-20)
>
>            mov     -#20,w0              ;temp0 =3D -20 =3D 0xFFEC
>            mov     w0,temp0
>            mov     -#32768,w1          ;temp1 =3D -32768 =3D 0x8000
>            mov     w1,temp1
>
>            mov     temp0,w0
>            mov     temp1,w1
>            sub     w0,w1,w2            ;w2 =3D 0x7FEC =3D -20 (bit15 =3D =
0)
>
> Questions -
>
> (1) what does - 32769 look like ?
> (2) what would be a method to test for < -32768 on a dsPIC ?
>
> Should I be looking at the 40-bit Accumulators for this ?
>
> TIA
>
> Joe
>  =20



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