Why does "int('1111', 2)" return "15" in Python?

Question

Why from this exit?

Code:

a = int('1111', 2)
print(a)

Exit:

15

Answer 1

You have created an integer based on 2, i.e., in binary, and 1111 is 15 in decimal.

1.2³ + 1.2² + 1.2¹ + 1.2⁰

8 + 4 + 2 + 1 = 15

If it were 1010 it would print 10. If it were 10000 it would print 16.

Documentation. Note that there is a first parameter that accepts the value that will be converted to integer and a second that says which base you want to use, the default is 10. So if you had used this base 1111 would print 1111 even.