If statement run code despite being false

Question

I'm very new to Python and to test my knowledge, I'm trying to build a Black Jack game. The code is rather long at this points, everything works fine, the following code is the only problem:

def player_round_end():
    global player_points
    if player_points_one == player_points_two:   
        # This works just fine
    elif player_points_one != player_points_two:
        print(f"---{player_points_one} or {player_points_two}")
        if player_points_one or player_points_two == 21: #This is the probelem
            print("---Black Jack!")

            if player_points_one == 21:
                player_points = player_points_one
            else:
                player_points = player_points_two

            print(f"Player points: {player_points}")
            dealer()      
        elif player_points_one and player_points_two > 21:
            print("You lost!")
            new_game()

I added some print functions so that I could easier find the bug. The problem with the == 21 statement is that it is sometimes defined as true, despite not beging true.

Example output from CMD:

---1 or 11
---Black Jack!
---Player points: 11

I drew and an A, which means that I have either 1 or 11 points. IT corectly defines that these are not equal, but it then defines that one of these is equal to 21, why is that? Let me know if you need more information or need to see all the code (though its not very pretty).

if player_points_one == 21 or player_points_two == 21? – drgs Jan 28 '19 at 12:43 — drgs, Jan 28 '19 at 12:43
Thank you, this was the problem! – Dennis Christiansen Jan 28 '19 at 12:46 — Dennis Christiansen, Jan 28 '19 at 12:46

score 2 · Accepted Answer · answered Jan 28 '19 at 12:45

2

if player_points_one or player_points_two == 21: ## Meaning player_points_one exists or player_points_two == 21

if player_points_one == 21 or player_points_two == 21: ## Correct solution

answered Jan 28 '19 at 12:45

Santhosh Kumar

543
8
22

1

Yeah, you are right, thank you! As far as I remember, the first statement would be correct in VBA but it obviously doesn't work in Python. – Dennis Christiansen Jan 28 '19 at 12:47
if its right, please feel free to select it as correct answer to close this question :) – Santhosh Kumar Jan 28 '19 at 12:48
1

I will!. But there is a 10 min timer from the post is created until I can pick the right solution :) – Dennis Christiansen Jan 28 '19 at 12:51
@DennisChristiansen why this answer got unaccepted if it worked – Santhosh Kumar Mar 06 '19 at 05:32
1

I don't know, I didn't do anything? As far as I remember I did give credit to you for being the first with the right answer? – Dennis Christiansen Mar 06 '19 at 06:25
may be a bug; because it showed my answer was unaccepted and down voted #StrangeBug – Santhosh Kumar Mar 06 '19 at 06:44

score 2 · Answer 2 · answered Jan 28 '19 at 12:50

2

player_points_one, being a non-zero number, is resolving to true.

This statement may make intuitive sense to a human reader:

if player_points_one or player_points_two == 21

But programming is a bit more explicit than human intuition. The conditions you're checking here are precisely:

player_points_one
player_points_two == 21

Whereas the conditions you want are:

player_points_one == 21
player_points_two == 21

Which would be:

if player_points_one == 21 or player_points_two == 21

answered Jan 28 '19 at 12:50

David

208,112
36
198
279

Thank you for the explanation, it makes sense! As I mentioned above, I'm completely new to Python, but in VBA you can actually do both, so I just assumed it worked the same way in Python. – Dennis Christiansen Jan 28 '19 at 12:59
1

@DennisChristiansen never assume that anything works the same way as VBA ;-) – bruno desthuilliers Jan 28 '19 at 13:59

If statement run code despite being false

2 Answers2