Automatic control function programming manual ladder language

4.5.8 Overflow - Sign Change
Literal Elements of Ladder Language
The system does not perform overflow control. It is therefore up to the programmer to take the necessary precautions.
Example of Sign Change:
%M100.B and %M101.B are two byte variables both equal to 0x7F (i.e. +127).
The assignment: %Var.xx = %M100.B + %M101.B gives:
0x00
0x00
0x00
Propagation of register bit 7
0x00
Propagation of register bit 7
0x00
0x00
%M100.B (0x7F)
+
%M101.B (0x7F)
= =
Var.xx
0 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1
Var.B is loaded with 0x00FE. Since bit 7 (sign bit) is a 1, Var.B == -2 (wrong answer) 1 1 1 1 1 1 1 0
Var.W is loaded with 0x00FE. Since bit 15 (sign bit) is a 0,
Var.W == 254 (right answer)
Var.L is loaded with 0x000000FE.
Since bit 31 (sign bit) is a 0,
Var.L == 254 (right answer)
0x00
4.5.9 Examples of Literal Entities
0x00
0x00
0x00
0x7F
0x7F
0xFE
0 0 0 0 0 0 0 0
0x00
Register D0
(on 32 bits)
Register D1
(on 32 bits)
D0 + D1 -> D0
Result D0
(on 32 bits)
0x00
1 1 1 1 1 1 1 0
Comparisons:
- %M5.B + %V33.L == %M10.W,
- (Var_1