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Why does `a == b or c or d` always evaluate to True?

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

import sys
print("Hello. Please enter your name:")
name = sys.stdin.readline().strip()
if name == "Kevin" or "Jon" or "Inbar":
print("Access granted.")
print("Access denied.")

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

Hello. Please enter your name:
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.


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 argument with a positive truth value:

if ("Jon"):

And since “Jon” has a positive truth value, 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 two 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 sequence of valid values, and use the in operator to test for membership:
    if name in {"Kevin", "Jon", "Inbar"}:

In general of the two 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'")
>>> timeit.timeit('name in {"Kevin", "Jon", "Inbar"}', setup="name='Inbar'")

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("if a == b or c or d or e: ...")
<_ast.Module object at 0x1031ae6a0>
>>> ast.dump(_)
"Module(body=[If(test=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())]), body=[Expr(value=Ellipsis())], orelse=[])])"

So the test of the if statement looks like this:

   left=Name(id='a', ctx=Load()),
   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 multiple values, namely, a == b and c, d, and e.

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