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?
I did come across that link but didn’t quite understand it. If looking only at 25 | 10, does the code not run as expected because 25 is 5 digits long and 10 is 4 digits long? Is that what’s meant by “two equivalent length bit designs”?
Also, I can’t tell if 10 | 4 = 7 or 10 | 4 = 14.
0d10 = 0b00001010 0d04 = 0b00000100 1010 | 0100 ------ 1110 0b00001110 = 0d14 0d25 = 0b00011001 0d10 = 0b00001010 11001 | 01010 ------- 11011 0b00011011 = 0d27If 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.
Thank you. 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 1111 is the binary representation of 7. This also has the equivalent result:
1 0 0 | 1 1 1 -------- 1 1 1All 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.