How to detect variable going below zero, without using PBPL ?

[CuriousOne]

So by checking the last bit, I can know, whenever variable went into negative world, right?

[Ioannis]

That is what post #3 says. Henrik explained it well.

Ioannis

[mpgmike]

Whether signed or unsigned, 0 = 0.
With an Unsigned BYTE (let's start simple), the range is 0 to 255. From 255, add 1 and you get 0 again, as the BYTE rolls over.
With a Signed BYTE, the range is -128 to +127. Think in terms of binary: 127d = %0111 1111. Make sense so far?

Working with a Signed BYTE, if you calculate: "0 - 1 =", you get %1111 1111 which equals -1.
With an Unsigned BYTE, if you calculate: "0 - 1 =" you still get %1111 1111, but it equals 255.

Think about this for Signed Variables:
0d = %0000 0000
1d = %0000 0001
-1d = %1111 1111
-2d = %1111 1110

Sometimes just looking at the representations enables the light bulb to turn on.

Now if we're working at the 16-bit WORD level:
0d = %0000 0000 0000 0000
1d = %0000 0000 0000 0001
-1d = %1111 1111 1111 1111
-2d = %1111 1111 1111 1110

See the trend?? I hope this helps you figure some things out.

[richard]

its called two's compliment , no magic involved

https://en.wikipedia.org/wiki/Two%27s_complement

[CuriousOne]

I came into another issue.
I have font array in EEPROM, and need to set offset for reading it. Depending on charset, this offset can be negative and positive.

So I have to do it in a way like this:

if langvar[x]=0 then
Y=(topline[x])*8+OFS

if langvar[x]=1 then
Y=(topline[x])*8-OFS

Can these two joined into single operation somehow?

[Ioannis]

if langvar[x]=0 then
    Y=(topline[x])*8+OFS
else
    Y=(topline[x])*8-OFS
endif

Ioannis

[CuriousOne]

No, I meant changing the value of OFS in the way, that addition will cause actual subtraction, due to overflow.