Bitwise operators and bit manipulation (no solution?)

Question

I've been racking my brain trying to come up with a solution, but am stuck.

I have a function, this function takes the input (either 0b0, 0b1, or 0b10) and sets the bit of some variable - or at least it's supposed to.

The value it's changing has two states: either 0b100 or 0b101. I want to set the respective bit from the input. This is easy for the true bits, but tricky for the false bits. Some pseudo code for all the scenarios:

if (var == 0b100 && input == 0b0) { do nothing }   // bit already set
if (var == 0b101 && input == 0b0) { var = 0b100 }  // bit is different, so we set it.

if (var == 0b100 && input == 0b1) { var = 0b101 }  // bit is different, so we set it.
if (var == 0b101 && input == 0b1) { do nothing }   // bit is already set

if (var == 0b100 && input == 0b10) { var = 0b110 } // bit is never set, so set it
if (var == 0b101 && input == 0b10) { var = 0b110 } // bit is never set, so set it

I don't want to cheat and make a separate statement for every possibility, I'd like a function for this. Here's what I've sort of strung together, but it obviously doesn't work:

if ( var ^ input )
{
    var ^= input;
}

The issue with this code is that 3rd bit (far left one) is always true.

Is this even possible?

EDIT:

Here is the same question, just posed a different way (which will result in different answers).

A function takes input for one bit. I want to take the left most bit and compare that bit and only that bit to another variable. e.g. input = 0b10 compared to 0b101's second bit, 0b00, since 0b10's left most bit is the second one.

On rare occasion, the unchanging bit may change :P (not from me though), so 0b001 or 0b000 may be it's value. For this reason, I cannot create a dynamic mask and compare only my bit.. The best method to solve my issue would be to answer this alternative question. Thanks a ton for your help, guys!

Answer 1

Step 1: Mask out the lower two bits.

var &= 0b100;

Step 2: Assign the lower two bits through OR operation.

var |= input;

Note that you may need to change the bit mask in step 1 in actual code base on the actual data type.