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COMMON QUESTIONS & ANSWERS
How are type conversions handled?
The compiler provides automatic type conversions when an assignment is performed. Some
information may be lost if the destination can not properly represent the source. For example:
int8var = int16var; Causes the top byte of int16var to be lost.
Assigning a smaller signed expression to a larger signed variable will result in the sign being
maintained. For example, a signed 8 bit int that is -1 when assigned to a 16 bit signed variable
is still -1.
Signed numbers that are negative when assigned to a unsigned number will cause the 2's
complement value to be assigned. For example, assigning -1 to a int8 will result in the int8 being
255. In this case the sign bit is not extended (conversion to unsigned is done before conversion
to more bits). This means the -1 assigned to a 16 bit unsigned is still 255.
Likewise assigning a large unsigned number to a signed variable of the same size or smaller will
result in the value being distorted. For example, assigning 255 to a signed int8 will result in -1.
The above assignment rules also apply to parameters passed to functions.
When a binary operator has operands of differing types then the lower order operand is
converted (using the above rules) to the higher. The order is as follows:
?
?
?
?
?
?
?
?
Float
Signed 32 bit
Unsigned 32 bit
Signed 16 bit
Unsigned 16 bit
Signed 8 bit
Unsigned 8 bit
1 bit
The result is then the same as the operands. Each operator in an expression is evaluated
independently. For example:
i32 = i16 - (i8 + i8)
The + operator is 8 bit, the result is converted to 16 bit after the addition and the - is 16 bit, that
result is converted to 32 bit and the assignment is done. Note that if i8 is 200 and i16 is 400
then the result in i32 is 256. (200 plus 200 is 144 with a 8 bit +)
Explicit conversion may be done at any point with (type) inserted before the expression to be
converted. For example in the above the perhaps desired effect may be achieved by doing:
i32 = i16 - ((long)i8 + i8)
In this case the first i8 is converted to 16 bit, then the add is a 16 bit add and the second i8 is
forced to 16 bit.
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