Why does the expression 0 < 0 == 0 return False in Python?

Question

Looking into Queue.py in Python 2.6, I found this construct that I found a bit strange:

def full(self):
    \"\"\"Return True if the queue is full, False oth         


        

Answer 1

maybe this excerpt from the docs can help:

These are the so-called “rich comparison” methods, and are called for comparison operators in preference to __cmp__() below. The correspondence between operator symbols and method names is as follows: x calls x.__lt__(y), x<=y calls x.__le__(y), x==y calls x.__eq__(y), x!=y and x<>y call x.__ne__(y), x>y calls x.__gt__(y), and x>=y calls x.__ge__(y).

A rich comparison method may return the singleton NotImplemented if it does not implement the operation for a given pair of arguments. By convention, False and True are returned for a successful comparison. However, these methods can return any value, so if the comparison operator is used in a Boolean context (e.g., in the condition of an if statement), Python will call bool() on the value to determine if the result is true or false.

There are no implied relationships among the comparison operators. The truth of x==y does not imply that x!=y is false. Accordingly, when defining __eq__(), one should also define __ne__() so that the operators will behave as expected. See the paragraph on __hash__() for some important notes on creating hashable objects which support custom comparison operations and are usable as dictionary keys.

There are no swapped-argument versions of these methods (to be used when the left argument does not support the operation but the right argument does); rather, __lt__() and __gt__() are each other’s reflection, __le__() and __ge__() are each other’s reflection, and __eq__() and __ne__() are their own reflection.

Arguments to rich comparison methods are never coerced.

These were comparisons but since you are chaining comparisons you should know that:

Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z, except that y is evaluated only once (but in both cases z is not evaluated at all when x < y is found to be false).

Formally, if a, b, c, ..., y, z are expressions and op1, op2, ..., opN are comparison operators, then a op1 b op2 c ... y opN z is equivalent to a op1 b and b op2 c and ... y opN z, except that each expression is evaluated at most once.