Why bitshift when evaluating truth tables as binary numbers?

Question

The last answer in this question shows that a binary truth table can be represented as a binary number:

0 0 0  | 1
0 0 1  | 1
0 1 0  | 0
0 1 1  | 0
1 0 0  | 1
1 0 1  | 0 
1 1 0  | 1
1 1 1  | 0

Can be represented by 01010011.

The entries in the table can also be evaluated using this number.

def evaluate(f, x):
    return (f & (1<<x)) != 0

f = int('01010011',2)

>>> evaluate(f,int('100',2))
True
>>> evaluate(f,int('101',2))
False

My question is about the evaluate function provided by the answer. Why must we left shift the bits by one?

Answer 1

You have it backwards. It is the binary number 1 shifted left by x spots.

And that should make sense. If you want to check the 4th spot in the table, represented by f, you have to check that f & 10000!=0.

You're right, shifting by one is very arbitrary and does not make sense.