Just a quick note to say your suggestions worked.

Thanks again. 

-----Original Message-----
From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On Behalf
Of Alex Parkinson
Sent: Tuesday, March 29, 2005 9:25 AM
To: Microcontroller discussion list - Public.
Subject: Re: [PIC]10-bit ADC integer formatting question

This is just speculation, but it sounds to me like the left shift
operation (ADRESH <<
8) is being performed on a single byte before being ORed and assigned to
VBattNew. 
Shifting a byte-sized variable by eight bits will shift all of the data
out of the variable, clearing it.  Try the following code:

VBattNew = ADRESH;
VBattNew = (VBattNew << 8) | ADRESL;

This should have the same result as your fix, without requiring a temp
variable.  It may be useful to examine the assembly code generated to
see exactly what is going on.

Regards,
Alex Parkinson


Leslie, Shawn J wrote:
> I am using the PIC18F4320. I am writing in C-language and using MPLAB
> C18 tool suite(compiler V2.2). I have a question about the ADC result 
> registers (ADRESL and ADRESH). I must use all 10 bits; therefore, I 
> have to do the simple task of taking the two 8-bit result registers 
> and combining them into a 16-bit integer. The following code is how I 
> did
> it:
> 
> (obviously before this I set the ADC result to be right justified and 
> VBattNew is defined as an unsigned int)
> 
> (I also send the raw values out the USART before the statements below 
> - getting exactly what one would expect in each of the bytes)
> 
> //Sent data out USART here
> VBattNew = ADRESL;		// Get low byte
> //Sent data out USART here
> VBattNew |= (ADRESH << 8);	// Combine with the high byte into 16
> int
> //Sent data out USART here
> 
> 
> The problem is the high byte continually gets cleared to 0x00 - I 
> cannot get past 255 decimal.
> 
> I even sent the raw data out the USART before, between and after for 
> inspection, and it is clear the left shift and integer assignment is 
> clearing the high byte?
> 
>  I tried the code below to make sure I was not trying to do much at 
> once with the above syntax - same problem:
> 
> //	VBattNew = ADRESL;			// Get low byte
> //	temp = ADRESH;
> //	temp << 8;
> //	VBattNew = temp | VBattNew;
> 
> The final fix which works is using the ADRES definition in the 
> p18f4320.h file - which apparently is the entire 16-bit concatenation 
> of the two bytes; as follows:
> 
> VBattNew = ADRES;	// This works
> 
> But, I want to understand why my code did not work - I bet it is 
> surprisingly simple (as always - its one bit screwing things up). By 
> the way I did search the archives and found someone posted code the 
> same way I did it - should be a slam dunk.
> 
> Thanks for your time guys,
> Shawn
> 
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