Bitwise '&' operator in chained comparison

问题

var = 86
print((var < 90) & (var >= 80))

prints True.

But why do all these print False?

print(var < 90 & var >= 80)
print(var < 90 & (var >= 80))
print((var < 90) & var >= 80)
print(var < 90 & True)

回答1:

You should be using the and operator instead for boolean operations.

Since python supports chaining relational operators (i.e. you can use 0 < var < 100 instead of 0 < var and var < 100) and processes binary operations (i.e. addition, subtraction, bitwise operations, etc.) before relational operators due to operator precedence, all of the failing cases you posted actually mean something else.

1. var < 90 & var >= 80 is equivalent to (var < 90 & var) and (90 & var >= 80)
2. var < 90 & (var>=80) is equivalent to var < 90 & True for which look at the 4th case.
3. (var<90) & var>=80 is also similar to the 4th case (this resolves to True & var >= 80 which will then resolve to 0 >= 80).
4. var < 90 & True is equivalent to var < (90 & True). The implementation of the & operator is designed to return 0 if either one of the operands is not an integer (which is another one of the biggest pitfalls of dynamic typing), which is why all such similar statements resolve to var < 0 which is false.

回答2:

It is because of Operator precedence

Take print( var<90 & True) as an example,

print( var<90 & True) is equivalent to print( var < (90 & True)) and hence it gives False.

However if you put in parenthesis in the right place, i.e. print( (var<90) & True), then it will give True.

回答3:

The bitwise operator & has higher precedence than the comparison operators, so the 90 & var is evaluated first, and then a chained comparison takes place.

>>> var = 86
>>> var<90 & var>=80 # var<90, and '90&var' is greater than or equal to 80
False
>>> var<(90 & var)>=80 # same as above
False
>>> (var<90) & (var>=80)
True

来源：https://stackoverflow.com/questions/41390988/bitwise-operator-in-chained-comparison