data type conversion - assignments between different lenghts

[Radomir]

Hello!

I have recently started using CCS compiler and I don't have much experience with C either.

I need to perform the following calculation:

var1*(var2/const), where:

var1 - 8bit integer
var2 - 8bit integer
const - 8 bit integer
1<var2<const

I assigned a 16bit temp var to hold var1*var2 and afterwards to divide it by const and assign it to the result var.
However, it seems that upper part of the temp var is lost.

Please, help!

[Ttelmah]

The key is that the arithmetic used, is defined by the sizes of the operands, not the 'target'. Hence if you multiply an int8, by an int8, the compiler will use int8 arithmetic.
You can force 16bit arithmetic to be used in a number of ways:
1) If you declare 'const' to be a long (for instance 100L), then this will force a switch to using int16 arithmetic
2) If you change one of the values into an int16, before performing the arithmetic, this will also force the switch. This is the 'preferred' way to go, and is standard C.
The arithmetic rules for the conversion, are exactly as written in K&R, but with the 'caveat', that the default 'int' type (which for instance a char will be converted to), is int8, rather than int16, which makes the conversion 'seem' less friendly.

So, use:

var1*((int16)var2/const)

Which will force var2, to be converted to an int16, before the arithmetic is started. Then the division will be done using an int16, and since the result is now an int16, the multiplication will also be done using the larger type.

Best Wishes

[Radomir]