Arithmetic expressions
When no parentheses are present, arithmetic expressions are evaluated from left to right, with multiplication and division having a higher priority than addition and subtraction.
The following table explains operator precedence in XC=BASIC:
Highest  NOT ,  (unary) 
* , / , MOD 

+ ,  

< , <= , > , = , >= , > 

Lowest  AND , OR , XOR 
The arithmetic operators, +
, 
, *
, and /
, work in the expected fashion, however, the result of arithmetic on type BYTE, WORD, INT or LONG cannot have a fractional result. Therefore 5 / 2
evaluates as 2, not 2.5. However, 5.0 / 2.0
evaluates as 2.5, because the presence of the decimal point tells the compiler that the numbers are of type FLOAT.
PRINT 5/2 ' Outputs 2 PRINT 5.0/2.0 ' Outputs 2.5
Operators, except unary operators, accept two operands. These two operands do not have to be of the same type. In a mixed expression, the operands are promoted to the higher type, being BYTE the lowest and FLOAT the highest type. The table below summarizes the results of a partially evaluated expression.
BYTE  WORD  INT  LONG  FLOAT  DECIMAL  

BYTE  BYTE  WORD  INT  LONG  FLOAT   
WORD  WORD  WORD  INT  LONG  FLOAT   
INT  INT  INT  INT  LONG  FLOAT   
LONG  LONG  LONG  LONG  LONG  FLOAT   
FLOAT  FLOAT  FLOAT  FLOAT  FLOAT  FLOAT   
DECIMAL            DECIMAL 
Note
DECIMAL types can not be used together with other types.
The type of the data being operated on must be considered. For example, adding two values of type BYTE will always result in a value which is also of type BYTE, even if the result is too large to fit in a BYTE. For example, if x
is a variable of type BYTE which has been previously assigned the value of 254, then the expression x + 4
will NOT have a value of 258, but 2. This is because BYTE variables can only take on values between 0 and 255, so that when you add 4 to 254, the result is (258256) = 2.
Likewise, adding or multiplying two numeric literals is subject to overflow, even if the left hand side of the assignment is a variable that could otherwise fit the result.
DIM a AS INT a = 250 + 6 PRINT a ' Outputs 0
Since both 250 and 6 are of type BYTE, the result will also be of type BYTE, regardless the type of a
. To overcome this problem, use the typecasting functions CBYTE, CWORD, CINT and CFLOAT.
DIM a AS INT a = CINT(250) + CINT(6) PRINT a ' Outputs 256