Byte operations

[RoboJeronimo]

Hi,

What are the result of...

DIM var AS BYTE 'max value=255 isn't it?
var=255
var=var+3
-------------------------------------------------------
DIM var AS BYTE 'min value=0 isn't it?
var=0
var=var-3

[NovaOne]

DIM var AS BYTE
var=255
var=var+3

So var = 2

DIM var AS BYTE
var=0
var=var-3

So var = 253

Is this wrong?

[RoboJeronimo]

So the secuencie is:

...253 - 254 - 255 - 0 - 1 - 2 - 3...

[TheJuggler]

That's all assuming that the byte is unsigned.

If it's signed then when you add 3 to 255, you'll get a negative number, and likewise when you subtract 3 from 0 you'll get -3. (although 255 isn't really possible since there are only 7 bits available for data meaning that you can get a max of 127.)

But if it's signed, then the maximum value is 127, as the most significant bit will be the sign bit.

[l3v3rz]

A byte is usually considered to be unsigned 8 bit number i.e. 0-255.

If you have an 8 bit signed integer the actual number range is -128 to 127.

[RoboJeronimo]

I check it again, with SERVO instruction.

var=255+3
SERVO 6, var

The variable is circular

[i-Bot]

Even though a BYTE variable, Robobasic first does the maths at 16 bits, so 255(0000 0000 1111 1111b) + 3(0000 0000 0000 00011b) = 258(0000 0001 0000 0010b). This is truncated to 8 bits when the variable is stored again as a BYTE, giving the value 2(0000 0010b).

For the subtract 0(0000 0000 0000 0000b) - 3(0000 0000 0000 0010) = 65533( C 1111 1111 1111 1101) note carry bit is set. This is truncated and carry bit ignored to give a BYTE of 253(1111 1101).

How did you check it ? the compiler may limit input values to 0 to 255, but the above is the way the maths is done when it is executed.