Reading port A

[misaeldv]

Hi, I have a question about reading port A, it is only 5 pins, which will be 5 bits (right). My question is if I read this into an int8, what would I get.
Will it be as follows?
Pin A1 is high,
Pin A2 is low
Pin A3 is high
Pin A4 is high
Pin A5 is low

int8 x;
x = input_a();

will
x = 00001101

First three zeros since there is only 5 bits, then it would go like A5, A4, A3, A2, A1?
or what will it be equal to?
Thank in advance

[misaeldv]

Sorry about that, it should be 6 bits for port A, A0 - A5;

int8 x;
x = input_a();

would x be as follows?
x = 00A5A4A3A2A1A0

Thanks.

[PCM programmer]

The safe way is to mask off the upper two bits. Then you know they
are set to 0 in 'x'.

[Ttelmah]

PCM_programmer has given you the safe answer. The other part of the answer (if you don't want to mask), is 'read the data sheet'. For example, on some of the chips, the high order bits of ports that don't have corresponding pins, are used internally. So (for instance), on the 12CE671/672/673/674, the hgh two bits, connect to the internal EEPROM (when available), and wake up reading '1', but may change if the eeprom is used. While on the 12F675, the memory data says that these bits are unimplemented and _will_ read as '0'.

Best Wishes

[misaeldv]

So, will this theoretically work in a switch statement? I tried it, but it seems to be doing random things.
Basically I'm not using A4 or A5, and A0 is being used as a switch, only A1 - A3 are being used to decide which action to perform, so what I am doing is the following, but it is not working; should this be working? Maybe there is something else going on somewhere in the program.

[PCM programmer]

[misaeldv]

Thanks, that makes a lot more sense, just one more question wouldn't shift right be the opposite?

[Ken Johnson]

I wouldn't shift it - just switch on the values you can get from the input.

Also, if you're looking at 3 bits, there are 8 possible combinations to check for. You may need to do something more complex, like remember the input from last time, and see which bits change.

Ken

[Ttelmah]

[quote="misaeldv"]Thanks, that makes a lot more sense, just one more question wouldn't shift right be the opposite?

[code]controller = (input_a() & 0x0E) <<1>> shifts a number right (towards the LSb). Effectively dividing it by two (if it is unsigned). << shifts a number left (towards the MSb), effectively doubling it (with the same coda about unsigned). For 'signed' values, the results are 'implementation specific' (varies according to the processor and compiler).
The arrows 'point' in the direction that the operation shifts.

Best Wishes