Alex - 1 year ago 71

Python Question

I have two numbers (binary or not, does not play any role) which differ in just one bit, e.g. (pseudocode)

`a = 11111111`

b = 11011111

I want a simple python function that returns the bit position that differs ('5' in the given example, when seen from right to left). My solution would be (python)

`math.log(abs(a-b))/math.log(2)`

but I wonder if there is a more elegant way to do this (without using floats etc.).

Thanks

Alex

Answer Source

You could use the binary exclusive:

```
a = 0b11111111
b = 0b11011111
diff = a^b # 0b100000
diff.bit_length()-1 # 5 (the first position (backwards) which differs, 0 if a==b )
```