Why does this piece of my program always trigger?

Question

Here is a excerpt from my program:

weaponinput = input("Would you like a rifle, a pistol, or a shotgun?")
if weaponinput == " pistol":
    weapon = (int(pistol_1))
if weaponinput == " rifle":
    weapon = (int(rifle_1))
if weaponinput == " shotgun":
    weapon = (int(shotgun_1))
if weaponinput != (" shotgun") or (" rifle") or (" pistol") or (" sniper rifle"):
    print("In your futile attempt to turn",weaponinput,"into a weapon you accidentally blow your brains accross the ground.")

The if clause always triggers in line 8, no matter the value of weaponinput. Why does this happen? I am using python, and do not really understand many other languages

You can't speak to a computer in human sentences. `weaponinput != "shotgun" or "rifle"` is equivalent to `(weaponinput != "shotgun") or "rifle"`. `"rifle"` is a nonempty collection, and evaluates as true. — Hammerite, May 08 '17 at 20:28
Are you actually typing a leading space when entering a value? You're checking for one, for some reason... — jasonharper, May 08 '17 at 20:28
Possible duplicate of [Why non-equality check of one variable against many values always returns true?](http://stackoverflow.com/questions/26337003/why-non-equality-check-of-one-variable-against-many-values-always-returns-true) — Barmar, May 08 '17 at 21:25

AShelly · Answer 1 · 2017-05-08T20:35:47.643

You have written the equivalent of

if (w != 1) or (2) or (3):
 print("something")

(2) is non-zero and therefore True. In your code ("rifle") is not None and therefore True.

The correct form is

if (w != 1) or (w!=2) or (w!=3):
    ...

Another way to do this might be

if weaponinput == "rifle:
    ...
elif weaponinput == "pistol": 
     ...
else:
    print("bad input message")

Yet another way:

WeaponCodes = {"pistol":int(pistol1), "rifle":int(rifle1), ... }
try:
   weapon = WeaponCodes[weaponinput]
except KeyError:
   print("bad input message")

score 2 · Answer 2 · answered May 08 '17 at 20:28

2

you need to change that line to the following:

if weaponinput != " shotgun" or weaponinput != " rifle" or weaponinput != " pistol" or weaponinput != " sniper rifle":

answered May 08 '17 at 20:28

Meccano

537
4
8

3

Or perhaps the more succinct `weaponinput not in [" shotgun", " rifle", " pistol", " sniper rifle"]` – Victor Chubukov May 08 '17 at 20:30
Thtat's plausible. – Meccano May 08 '17 at 20:31

score 1 · Accepted Answer · answered May 08 '17 at 20:31

1

Pythonic would be:

if weaponinput not in (" shotgun", " rifle", " pistol", " sniper rifle"):
    print(...)

answered May 08 '17 at 20:31

Maurice Meyer

17,279
4
30
47

Thanks. This will help me in my other projects too. – Anonymoose May 08 '17 at 23:37

score 0 · Answer 4 · answered May 08 '17 at 20:35

To check multiple conditions you have to compare your input variable with the required condition like:

if var <conditional operator1> condition1 <logical operator> var <conditional operator2> condition2 and so on.

so in your case line number 8 must be: if weaponinput != " shotgun" or weaponinput != " rifle" or weaponinput != " pistol" or weaponinput != " sniper rifle":

score 0 · Answer 5 · answered May 08 '17 at 20:43

Out of all operators, the logical operators(or, and and not) has the least priority (Python Operator Precedence).

In Python the Boolean value of a non-empty string is always true.

So, the code you have written is equivalent to :-

weaponinput = input("Would you like a rifle, a pistol, or a shotgun?")
if weaponinput == " pistol":
    weapon = (int(pistol_1))
if weaponinput == " rifle":
    weapon = (int(rifle_1))
if weaponinput == " shotgun":
    weapon = (int(shotgun_1))
if weaponinput != (" shotgun") or True or True or True:
    print("In your futile attempt to turn",weaponinput,"into a weapon you accidentally blow your brains accross the ground.")

So, the condition in line 8 is always true irrespective of whether weaponinput is equal to " shotgun" or not(as or always evaluates to true if atleast one of its operand is true).

Why does this piece of my program always trigger?

5 Answers5