181

I am writing a security system that denies access to unauthorized users.

name = input("Hello. Please enter your name: ")
if name == "Kevin" or "Jon" or "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

It grants access to authorized users as expected, but it also lets in unauthorized users!

Hello. Please enter your name: Bob
Access granted.

Why does this occur? I've plainly stated to only grant access when name equals Kevin, Jon, or Inbar. I have also tried the opposite logic, if "Kevin" or "Jon" or "Inbar" == name, but the result is the same.


This question is intended as the canonical duplicate target of this very common problem. There is another popular question How to test multiple variables for equality against a single value? that has the same fundamental problem, but the comparison targets are reversed. This question should not be closed as a duplicate of that one as this problem is encountered by newcomers to Python who might have difficulties applying the knowledge from the reversed question to their problem.

For in instead of ==, there are solutions here: How to test the membership of multiple values in a list

2
  • 17
    Variations of this problem include x or y in z, x and y in z, x != y and z and a few others. While not exactly identical to this question, the root cause is the same for all of them. Just wanted to point that out in case anyone got their question closed as duplicate of this and wasn't sure how it's relevant to them.
    – Aran-Fey
    Commented Apr 11, 2019 at 8:55
  • 1
    I disagree with the argument that this should not be closed as a duplicate. The comparison may be reversed, but that is irrelevant--the problem and the solution(s) are the same. As the post's revision history shows, many users (including gold badge holders and diamond moderators) seem to agree.
    – TylerH
    Commented Mar 25 at 18:08

8 Answers 8

233

In many cases, Python looks and behaves like natural English, but this is one case where that abstraction fails. People can use context clues to determine that "Jon" and "Inbar" are objects joined to the verb "equals", but the Python interpreter is more literal minded.

if name == "Kevin" or "Jon" or "Inbar":

is logically equivalent to:

if (name == "Kevin") or ("Jon") or ("Inbar"):

Which, for user Bob, is equivalent to:

if (False) or ("Jon") or ("Inbar"):

The or operator chooses the first operand that is "truthy", i.e. which would satisfy an if condition (or the last one, if none of them are "truthy"):

if "Jon":

Since "Jon" is truthy, the if block executes. That is what causes "Access granted" to be printed regardless of the name given.

All of this reasoning also applies to the expression if "Kevin" or "Jon" or "Inbar" == name. the first value, "Kevin", is true, so the if block executes.


There are three common ways to properly construct this conditional.

  1. Use multiple == operators to explicitly check against each value:

    if name == "Kevin" or name == "Jon" or name == "Inbar":
    
  2. Compose a collection of valid values (a set, a list or a tuple for example), and use the in operator to test for membership:

    if name in {"Kevin", "Jon", "Inbar"}:
    
  3. Use any() and a generator expression to explicitly check against each value in a loop:

    if any(name == auth for auth in ["Kevin", "Jon", "Inbar"]):
    

In general the second should be preferred as it's easier to read and also faster:

>>> import timeit
>>> timeit.timeit('name == "Kevin" or name == "Jon" or name == "Inbar"',
    setup="name='Inbar'")
0.0960568820592016
>>> timeit.timeit('name in {"Kevin", "Jon", "Inbar"}', setup="name='Inbar'")
0.034957461059093475
>>> timeit.timeit('any(name == auth for auth in ["Kevin", "Jon", "Inbar"])',
    setup="name='Inbar'")
0.6511583919636905

For those who may want proof that if a == b or c or d or e: ... is indeed parsed like this. The built-in ast module provides an answer:

>>> import ast
>>> ast.parse("a == b or c or d or e", "<string>", "eval")
<ast.Expression object at 0x7f929c898220>
>>> print(ast.dump(_, indent=4))
Expression(
    body=BoolOp(
        op=Or(),
        values=[
            Compare(
                left=Name(id='a', ctx=Load()),
                ops=[
                    Eq()],
                comparators=[
                    Name(id='b', ctx=Load())]),
            Name(id='c', ctx=Load()),
            Name(id='d', ctx=Load()),
            Name(id='e', ctx=Load())]))

As one can see, it's the boolean operator or applied to four sub-expressions: comparison a == b; and simple expressions c, d, and e.

6
  • 1
    Is there a specific reason to choose a tuple ("Kevin", "Jon", "Inbar") instead of a set {"Kevin", "Jon", "Inbar"} ?
    – Human
    Commented Jun 14, 2019 at 12:04
  • 3
    Not really, since both work if the values are all hashable. Set membership testing has better big-O complexity than tuple membership testing, but constructing a set is a little more expensive than constructing a tuple. I think it's largely a wash for small collections like these. Playing around with timeit, a in {b, c, d} is about twice as fast as a in (b, c, d) on my machine. Something to think about if this is a performance-critical piece of code.
    – Kevin
    Commented Jun 14, 2019 at 12:13
  • 3
    Tuple or list when using 'in' in an 'if' clause? recommends set literals for membership testing. I'll update my post.
    – Kevin
    Commented Jun 14, 2019 at 12:20
  • 4
    In modern Python, it recognizes that the set is a constant and makes it a frozenset instead, so the constructing set overhead is not there. dis.dis(compile("1 in {1, 2, 3}", '<stdin>', 'eval'))
    – endolith
    Commented May 16, 2020 at 17:48
  • FWIW I think you should re-add the tuple as that is simpler for folks to understand than a set.
    – TylerH
    Commented Oct 24, 2022 at 13:54
8

Summarising all existing answers

(And adding a few of my points)

Explanation :

if name == "Kevin" or "Jon" or "Inbar":

is logically equivalent to:

if (name == "Kevin") or ("Jon") or ("Inbar"):

Which, for user Bob, is equivalent to:

if (False) or ("Jon") or ("Inbar"):

NOTE : Python evaluates the logical value of any non-zero integer as True. Therefore, all Non-empty lists, sets, strings, etc. are evaluable and return True

The or operator chooses the first argument with a positive truth value.

Therefore, "Jon" has a positive truth value and the if block executes, since it is now equivalent to

if (False) or (True) or (True):

That is what causes "Access granted" to be printed regardless of the name input.

Solutions :

Solution 1 : Use multiple == operators to explicitly check against each value

if name == "Kevin" or name == "Jon" or name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Solution 2 : Compose a collection of valid values (a set, a list or a tuple for example), and use the in operator to test for membership (faster, preferred method)

if name in {"Kevin", "Jon", "Inbar"}:
    print("Access granted.")
else:
    print("Access denied.")

OR

if name in ["Kevin", "Jon", "Inbar"]:
    print("Access granted.")
else:
    print("Access denied.")

Solution 3 : Use the basic (and not very efficient) if-elif-else structure

if name == "Kevin":
    print("Access granted.")
elif name == "Jon":
    print("Access granted.")
elif name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")
6

There are 3 condition checks in if name == "Kevin" or "Jon" or "Inbar":

  • name == "Kevin"
  • "Jon"
  • "Inbar"

and this if statement is equivalent to

if name == "Kevin":
    print("Access granted.")
elif "Jon":
    print("Access granted.")
elif "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Since elif "Jon" will always be true so access to any user is granted

Solution


You can use any one method below

Fast

if name in ["Kevin", "Jon", "Inbar"]:
    print("Access granted.")
else:
    print("Access denied.")

Slow

if name == "Kevin" or name == "Jon" or name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Slow + Unnecessary code

if name == "Kevin":
    print("Access granted.")
elif name == "Jon":
    print("Access granted.")
elif name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")
0
3

Not-empty lists, sets, strings, etc. are evaluable and, therefore, return True.

Therefore, when you say:

a = "Raul"
if a == "Kevin" or "John" or "Inbar":
    pass

You are actually saying:

if "Raul" == "Kevin" or "John" != "" or "Inbar" != "":
    pass

Since at least one of "John" and "Inbar" is not an empty string, the whole expression always returns True!

The solution:

a = "Raul"
if a == "Kevin" or a == "John" or a == "Inbar":
    pass

or:

a = "Raul"
if a in {"Kevin", "John", "Inbar"}:
    pass
1
  • 2
    good otherwise but "You are actually saying:" is wrong, that's not how or works. The value of the expression is "John", not True. Commented Apr 11, 2021 at 18:50
3

Using match/case in Python 3.10 and above

Python 3.10 adds a new syntax to the language. It's officially described as "structural pattern matching", but most people call it according to the syntax: "match/case".

We can use this special syntax for an example like in the question, by making one "case" that matches all the accepted usernames, and using the "wildcard" case _ in place of the else. Thus:

name = input("Hello. Please enter your name: ")
match name:
    case "Kevin" | "Jon" | "Inbar":
        print("Access granted.")
    case _:
        print("Access denied.")

Note that cases are "combined" using |, not or. This is a special syntax: Python does not try to compute "Kevin" | "Jon" | "Inbar" first (| doesn't work with strings), but instead interprets the entire line differently because it starts with case.

2

Simple engineering problem, let's simply it a bit further.

In [1]: a,b,c,d=1,2,3,4
In [2]: a==b
Out[2]: False

But, inherited from the language C, Python evaluates the logical value of a non zero integer as True.

In [11]: if 3:
    ...:     print ("yey")
    ...:
yey

Now, Python builds on that logic and let you use logic literals such as or on integers, and so

In [9]: False or 3
Out[9]: 3

Finally

In [4]: a==b or c or d
Out[4]: 3

The proper way to write it would be:

In [13]: if a in (b,c,d):
    ...:     print('Access granted')

For safety I'd also suggest you don't hard code passwords.

0

Besides some other rather rarer useful cases for the walrus operator already mentioned. This also tend to be a useful case as well.

def calc_value():
    return 43

if (v := calc_value()) == 43 and v > 42:
    print('happy short, efficient and readable code')

This works because each part of the if-statement is read separately. So (v := calc_value()) is executed and a value is assigned to v and if the first fails, you still have v in the namespace for different conditions or calculations.

-4

Approaches

How a data scientist approaches this problem

The simplest way possible is to eliminate the need for comparison operators and use a list. This looks impressive on security systems because you learn to access ORMs.

user = input("Enter name: ")

if user in {"Bob", "Kevin", "Joe"}:
   print("Access granted, " + str(user) + ".")
else:
   print("Access denied.")

Or, you can resemble the exact same code above, just put the list of registered users in their own list:

user = input("Enter name: ")
users = {"Bob", "Kevin", "Joe", "a million more users if you like"}

if user in users:
   print("Access granted, " + str(user) + ".")
else:
   print("Access denied.")

If you wanted to complete this protocol safely without the risk of attack, set up double parameters. This would check your mini-ORM for first and last name fields, as well as a password or secret question key. Objects can be sorted like this if you want to efficiently lazy-load user credentials without hashing:

def lazy(i):
   j = 0 # For example
   while j < i:
      yield j
      j += 1

The loop will consume only the yielded values to save time and energy on your system:

You can then do something with the iterated list:

for j in lazy_range(10):
   do_something_here(j)

This problem can be approached from any angle: memory management, security, or simply by an organic list or packaged ORM.

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