if coin == 25 | 10 | 5:

If I replace the ‘|’ with ‘or’ the code runs just fine. I’m not sure why I can’t use ‘|’ in the same statement.

Doing the following doesn’t work either:

if coin == 25 | coin == 10 | coin == 5:

I know bitwise operators can only be used with integers, but other then that is there another difference from logical operators?

  • Lucy :3@feddit.org
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    20 days ago
    0d10 = 0b00001010
    0d04 = 0b00000100
    
    1010   |
    0100
    ------
    1110
    
    0b00001110 = 0d14
    
    
    0d25 = 0b00011001
    0d10 = 0b00001010
    
    11001   |
    01010
    -------
    11011
    
    0b00011011 = 0d27
    

    If an int is only x bytes long, but y bytes are needed for an action, it’s just padded with 0’s to the left. So 0b110 = 0b0110 = 0b00000110. Usually you will use either 8, 16, 32 or 64 digits/bits (char, short, int and long respectively, and 1, 2, 4 and 8 bytes long, 1 byte = 8 bits. So you’ll usually align it with bytes.) However, for calculating like this removing trailing zeroes can make it more tidy, and sometimes 4 or 2 bits are used too. And bools technically only use 1 bit.

    • milon@lemm.eeOP
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      20 days ago

      Thank you. Not sure why in the link the arithmetic in green results in 7.

      • sugar_in_your_tea@sh.itjust.works
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        8 days ago

        Not sure why in the link the arithmetic in green results in 7.

        Here’s the example from the link:

          1 0 0
        | 0 1 1
        --------
          1 1 1
        

        111 is the binary representation of 7. This also has the equivalent result:

          1 0 0
        | 1 1 1
        --------
          1 1 1
        

        All it’s doing is going through each column of the binary number, and if at least one of them is a 1, the result is a 1. If they’re both a 0, the result is a zero. Then you just convert from binary to decimal, so 111 -> 7.