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How does Python parse the first of the following three expressions? (I expected it to be the same as the second, since == and in have the same precedence.)
>>> 1 == 0 in (0,1), (1==0) in (0,1), 1 == (0 in (0,1))
(False, True, True)
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False == (False or True) In this expression python would first solve the comparison operator in bracket. => (False or True) is said to be True. Because the 'OR' operator search for the first truthy value, and if found it returns the value else 'False'.
Zero is used to represent false, and One is used to represent true. For interpretation, Zero is interpreted as false and anything non-zero is interpreted as true. To make life easier, C Programmers typically define the terms "true" and "false" to have values 1 and 0 respectively.
Because they don't represent equally convertible types/values. The conversion used by == is much more complex than a simple toBoolean conversion used by if ('true') . So given this code true == 'true' , it finds this: "If Type(x) is Boolean , return the result of the comparison ToNumber(x) == y ."
Like in C, the integers 0 (false) and 1 (true—in fact any nonzero integer) are used.
See the documentation of comparison operators: they are chained rather than grouped. So 1 == 0 in (0,1) is equivalent to (1==0) and (0 in (0,1)), which is obviously false.
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