10 bit ADC basics

[rovtech]

Question 1.
Is the default actually 8 bit?
I am using a PIC16F1509 and my program works reading the ADC into an 8 bit variable.

Question 2.
Should I be using #device ADC=8 or is using the default OK (if there is one)?

Question 3.
After searching for answers I notice #device *=16 ADC=10 being used. What exactly is *=16 ? The CCS manual suggests it sets the pointer size. What pointer and why?

Question 4.
What is the default justification for 10 bits returned in a 16 bit variable?

Question 5.
How would I change a 16 bit value into an 8 bit variable?

Thanks in advance.

[Ttelmah]

ADC=8, is the default, but it is always safer to specify it yourself.

ADC=10, or ADC=16, take longer to read, larger code, and slower. If you only want an 8bit value, stick with ADC=8.

ADC=10 returns the value right justified (low ten bits). ADC=16 returns the value left justified (to the top of the 16bit result). 10bit value to 8bits, you'd have to divide by 4. To just read the top byte or low byte of a value look at the 'make8' instruction.

*=16, is not connected with the ADC, and can even be on another line. When you 'address' values in RAM, the address is called a 'pointer' (this is the C name for the address). The RAM memory in the PIC16 is 'paged'. So to talk to things bigger than one 'page' in RAM, a multi byte address is needed, and code has to be added to automatically select this. The default is to only handle the RAM in the bottom page of memory (quick). *=16, tells the compiler to generate code to address the whole of the RAM, allowing a maximum of 64KB of RAM to be addressed (doesn't do anything on larger chips without the RAM paging PIC18/24 etc. - often misused on these)

Best Wishes

[rovtech]

That was fast and informative, Thanks.

[rovtech]

Here are two examples and some more questions. Only relevant code is shown.
// Example using an 8 bit variable and reading a 10 bit ADC as 8 bit