I have this simple line of code:
i = " "
if i != "" or i != " ":
print("Something")
This should be simple, if i is not empty ""
OR it's not a space " "
, but it is, print Something. Now, why I see Something printed if one of those 2 conditions is False
?
De Morgan's laws,
"not (A and B)" is the same as "(not A) or (not B)"
also,
"not (A or B)" is the same as "(not A) and (not B)".
In your case, as per the first statement, you have effectively written
if not (i == "" and i == " "):
which is not possible to occur. So whatever may be the input, (i == "" and i == " ")
will always return False
and negating it will give True
always.
Instead, you should have written it like this
if i != "" and i != " ":
or as per the quoted second statement from the De Morgan's law,
if not (i == "" or i == " "):
This condition:
if i != "" or i != " ":
will always be true. You probably want and
instead of or
...
I will explain how or
works.
If checks the first condition and if it is true it does not even check for the second condition.
If the first condition is false only then it checks for second condition and if it is true the whole thing becomes true.
Because
A B Result
0 0 0
0 1 1
1 0 1
1 1 1
So If you want to satisfy both condition of not empty and space use and
Your print statement will always happen, because your logic statement is always going to be True.
if A or B:
will be True if either A is True OR B is True OR both are True. Because of the way you've written the statement, one of the two will always be True. More precisely, with your statement as written, the if statement correlates to if True or False:
which simplifies to if True:
.
It seems that you want an and
statement instead of an or
.
i = " "
you have the condition as
if i != "" or i != " ":
here i != ""
will evaluate to True
and i != " "
will evaluate to False
so you will have True or False
= True
you can refer this truth table for OR
here
True or False = True
False or True = True
True or True = True
False or False = False