ATMega8 - PORT C - Digital Output Issue
Your problem:

Your DDR setting is incorrect: You've written

DDRC |= _BV(PC0) | _BV(PC1) | _BV(PC3) &amp; _BV(PC2);


but you want 

DDRC |= _BV(PC0) | _BV(PC1) | _BV(PC3)
DDRC &amp;= ~_BV(PC2);


You say you want more detail because you just kinda emulated other DDRC assignments and don't really understand what you wrote?  OK, here goes: 

How to set registers with bitwise operators on the AVR.

Let's unpack the _BV() macro and PCx definitions for a moment.  _BV(x) is simply 1 &lt;&lt; x.  If you write _BV(5), the macro will evaluate to the number 0b 0001 0000.  PCx is simply a macro defined to the value of x, so _BV(PC5) will evaluate to the same thing as _BV(5).   You're probably expecting something like the following to happen: 

   0b 0000 0001  // PC0
|  0b 0000 0010  // PC1
|  0b 0000 1000  // PC3
&amp;  0b 0000 0100  // Misconception: AND does not cause this to output a zero at the set bit
|= ------------
   0b 0000 1011


However, you're not clearing PC2 correctly.  You're applying a binary AND at the end.  AND has a higher precedence than OR in C, so your operation is actually 

DDRC |= _BV(PC0) | _BV(PC1) | ( _BV(PC3) &amp; _BV(PC2) );


PC3 is 0b 0000 1000, and PC2 is 0b 0000 0100, so the parenthetical section evaluates to zero.  This is actually fortunate for PC2, but unfortunate for PC3, which remains clear when you wanted it set. To clear a bit, you do use &amp;, but you need to take the complement of the number using the ~ operator.  In the following equations, I've used ? for bits which we don't want to change:  

   0b ???? ???1  // Set PC0
|  0b ???? ??10  // Set PC1 
|  0b ???? 1?00  // Set PC3
|= ------------
   0b ???? 1?11

   0b ???? 1?11
&amp;  0b 1111 1011  // Clear PC2
&amp;= ------------
   0b ???? 1011


In code, these equations are: 

DDRC |= _BV(PC0) | _BV(PC1) | _BV(PC3)
DDRC &amp;= ~_BV(PC2);


as shown above.  You could either do this in two steps as I did, or apply the previous value of DDRC with a bit mask: 

DDRC = (DDRC &amp; 0xF0) | ( ( _BV(PC0) | _BV(PC1) | _BV(PC3) ) &amp; ~_BV(PC2) );


0xF0, or 0b 1111 0000, when AND'ed with the previous value of DDRC, returns 0b ???? 0000, to which we can OR our lower four bits.  I think the two-step method is clearer: Your inputs and outputs are on different lines (note that you can add additional inputs with, for example, DDRC &amp;= ~_BV(PC2) &amp; ~_BV(PC4)).  Sometimes, I'll even break it up onto more lines: 

DDRC |= _BV(PC0)   // RELAY STATUS LED - O/P
DDRC |= _BV(PC1)   // RELAY CTL LINE - O/P
DDRC &amp;= ~_BV(PC2)  // PUSH BUTTON - I/P
DDRC |= _BV(PC3)   // SPARE - O/P


That, to me, is the most clear mechanism possible, and the easiest for future readers to understand.  It's also the most verbose, but who cares?  The compiler will optimize all three mechanisms to the same code.