When doing ACSR |=(1<<ACD)|(1<<ACBG) are the rest of the bits assumed to be 0?

[jerive]

When doing ACSR |=(1<<ACD)|(1<<ACBG) are the rest of the bits (on the right side of the equation) assumed to be 0? or do they need to be explicitly defined?

[DrDiettrich]

The 1 on the right hand side has exactly one bit set. Consequently all others are zero.

[johnwasser]

@jerive

ACSR |= (1<<ACD) | (1<<ACBG);

You are setting those two bits and leaving the others unchanged. If you want to set those two and clear the rest:
ACSR = (1<<ACD) | (1<<ACBG);

Note: In the Analog Compare Status Register, Bit 5 (ACO) is read-only. It doesn't matter what you try to write to it.

Note: In the Analog Compare Status Register, Bit 4 (ACI) is an interrupt flag. It gets CLEARED if you write a 1 to it.