Boolean testing: Python prints '1' or 'True'

Question

I did a simple test like below:

>>> valsH["S0"] = 1
>>> valsH["I1"] = 0
>>> valsH["I2"] = 1
>>> valsH["I0"] = 1
>>> """original position of: not valsH["I1"]"""
>>>
>>> valsH["I0"] and not valsH["I1"] and valsH["I2"] and valsH["S0"]
1
>>> """After moving: not valsH["I1"] to the end of the expression"""
>>>
>>> valsH["I0"] and valsH["I2"] and valsH["S0"] and not valsH["I1"]
True
>>> 

So it seems that depending on where

not valsH["I1"]

is, the value of the Boolean equation is printed as '1' or 'True'.

Why is this so?

Answer 1

This is by how and works. Let's take the following example:

p and q

First evaluate p, if p is true, evaluate q. If q is true, it returns True but of whatever type q. So:

1 and True # True
True and 1 # 1 

For or it goes the other way around. Since only one is required to be true, if the first is true, it returns true of that type. So:

1 or True # 1
True or 1 # True

Answer 2

When you use logical and operation Python evaluates the last operand enough to give an answer. For instance:

If it is :

1 and True and 1

It will return 1 because it needs to go till the end (all is True) and it will return what is in the end, i.e. 1.

Or:

1 and 1 and True

It will return True because True is the last operand.

Another case:

1 and 0 and False

It will return 0 because it stopped evaluation on 0 which results in 0.

False and 0 and 1

Will return False because it stopped evaluation on False which results in False.

Hope this helps.