Brazilian greetings!
First of all thanks Joby for your example. Secondly, his example has just a minor error. The number 0x20 is not correct. It should be 0x04. Also, just as a suggestion, I would not use hexadecimal numbers like 0xFB, 0x20, or 0x04 in the code. I would suggest using the PIN port definitions found in the io.h and other ones referenced by header file. I have rewritten Joby's example below, with some comments for the beginners.
# include <avr/io.h>
int main (void)
{
// set all pins on PORTB for output
DDRB = 0xFF;
// set port pin PORTD2 as input and leave the others pins
// in their originally state (inputs or outputs, it doesn't matter)
DDRD &= ~(1 << PD2); // see comment #1
while (1)
{
if (PIND & (1<<PD2)) // see comment #2
PORTB |= (1<<PB2); // see comment #3
else
PORTB &= ~(1<<PB2); // see comment #4
}
return 0;
}
/*
comments for beginners
comment #1: (1 << PD2) generates the binary 00000100.
The operation "~" flips all the digits, i.e.,
the binary now is 11111011. Finally the &=
applies the logic "AND" between DDRD and 11111011
and the result is placed again in DDRD memory.
Note: What the operator "AND" does is for each bit
in the DDRD memory, it compares with the binary number
above. If the bit in DDRD is 0 and the bit in the
binary at the same bite position is 1, then the
resulting bit is 0, if the DDRD is 1 and the
bit in the binary is 1, the resulting bit is 1,
and if the bit in the DDRD is 1 or 0 and
the bit in the binary is 0 then the resulting
bit is always 0. In summary, the command DDRD &= ~(1 << PD2)
changes only the bit PD2 to zero and leave the other ones
(zeros or ones) untouched. It seems a little
bit complicated, but after you get used to it, it
is the best way of changing a bit in a bite without
changing the other bits.
comment #2: (1 << PD2) generates the binary 00000100.
Using the same logic "AND" described in
comment #1, the command "PIND & 0000100" checks
only if the PIND2 (our input pin where the push button
is connected to) is set to high or not. All the other
pins will be FALSE since the binary bits are set to 0,
and since the binary bit #2 is set to 1, the IF statement
will be TRUE only if the PD2 input is set to high
or FALSE if the PD2 input is set to low.
comment #3: Following the logic explained in comment #1, this
command sets output pin PINB2 in port PORTB to
high voltage. If your LED is correct connected to
this pin port with a resistor of ~300 ohms, and that
resistor is connected to the ground, the LED should turn on.
comment #4: The LED should turn off for the same reasons explained
in the previous comments.
Final considerations:
a) To avoid voltage oscillation in the input pin PD2 when the
push button is not pressed (open circuit), I strong recommend
to place a pull-down resistor (1 kOhm or higher), so that the
LED does not light up accidentally due to this random voltage
oscillation.
b) A disclaim note: The ideas described here are to be used as
educational only and they should NOT be used in any real system
before consulting an expert in electronics.
*/
