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So I am on pset6 trying to implement the credit card check.

I've already completed working through the numbers and am not having any issue with the actual algorithm.

But for some reason my if statements logic doesn't work.

if ((foo)%10 == 0):
    a = True
else:
    a = False

print(r[0])
print(r[1])

if (a == True and r[0] == 4):
    print("VISA")
elif(a == True and r[0] == 3 and r[1] == 4 or 7):
    print("AMEX")
elif(a == True and r[0] == 5 and r[1] == 1 or 2 or 3 or 4 or 5):
    print("MASTERCARD")

See here

Even though I've set the logic up to only recognize as a Amex if r[0] is 3 and r[1] is 4 or 7. It still shows up as an AMEX. I have also printed out r[0] and r[1] so I can show you that the values within that index are correct.

I don't understand how this can ignore my logic statements and print anyway. Please help me understand what I have done incorrectly. Thank you for your time.

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  • Am I not providing enough information or something? I'd really like to figure out what I am doing wrong here.
    – Multiplify
    Commented May 3, 2021 at 13:16

1 Answer 1

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This elif(a == True and r[0] == 3 and r[1] == 4 or 7): will always evaluate to True because 7 is "truthy". Did you mean or r[1] == 7? And once an or introduced, the expression will likely need parentheses. Same goes for the subsequent elif.

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  • Hmm, I'm not sure I understand what you mean by 7 evaluating as truthy. So should I re-write the code as r[1] == 4 or ==7 or do I need to write r[1] == 4 or r[1] == 7? I'll go and test it now. When we are writing or statements do we need to re-iterate the element every time? Is there a better way for me to write if r[1] is within 1-5 do something
    – Multiplify
    Commented May 3, 2021 at 19:15

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