Thanks, I was looking for this.

Stefano


Jan van Wijk ha scritto:
> Hello Stefano,
> 
> On Fri, 30 Jan 2009 08:48:31 +0100, Stefano Vanzo wrote:
> 
>>>> With 16bit models can I do:
>>>>
>>>> DATA_PORT |= data8bit & 0x00f0;
> 
> Yes, allthough I do not understand why you 'and' with 0x00f0.
> That will ZERO the four lowest bits from your data8bit value.
> But that is probably part of yor application logic :-)
> 
> 
>>>> or do I have to:
>>>>
>>>> int data16bit = 0x0000;
>>>> data16bit += data8bit;
>>>> DATA_PORT |= data16bit & 0x00f0;
> 
> No, that has the same effect as the above ...
> The 8-bit value is automatically promoted to 16 bit.
> 
>>>> The question: can I "and" a char with a int ending up with the char in
>>>> the first two nibbles?
> 
> Yes 
> 
> <snip>
>> I have two 16bits because DATA_PORT is 16 bit. I want to know if 
>> data8bit ends up in the lowwer part of DATA_PORT and does not effect the 
>> two high nibbles.
> 
> Well, it DOES affect the high nibbles in principle, but since your operation
> is an OR, and the two high-nibbles in your argument are ZERO, no change
> will take place. (the |= causes a 16-bit read-modify-write)
> 
> Your 8-bit value is promoted to a 16-bit value before that operation,
> in this case by adding two zero nibbles.
>> DATA_PORT |= data8bit & 0x00f0;
> 
> For readability, I would add parentheses on the 'and' clause:
> 
> DATA_PORT |= (data8bit & 0x00f0);
> 
> Regards, JvW
> 
> Jan van Wijk, author of DFSee; http://www.dfsee.com
> -------------------------------------------------------------------------------

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