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

Question

I believe Python has special case handling for sequences of relational operators to make range comparisons easy to express. It's much nicer to be able to say 0 < x <= 5 than to say (0 < x) and (x <= 5).

These are called chained comparisons. And that's a link to the documentation for them.

With the other cases you talk about, the parenthesis force one relational operator to be applied before the other, and so they are no longer chained comparisons. And since True and False have values as integers you get the answers you do out of the parenthesized versions.

Because

(0 < 0) and (0 == 0)

is False. You can chain together comparison operators and they are automatically expanded out into the pairwise comparisons.

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EDIT -- clarification about True and False in Python

In Python True and False are just instances of bool, which is a subclass of int. In other words, True really is just 1.

The point of this is that you can use the result of a boolean comparison exactly like an integer. This leads to confusing things like

>>> (1==1)+(1==1)
2
>>> (2<1)<1
True

But these will only happen if you parenthesise the comparisons so that they are evaluated first. Otherwise Python will expand out the comparison operators.

The strange behavior your experiencing comes from pythons ability to chain conditions. Since it finds 0 is not less than 0, it decides the entire expression evaluates to false. As soon as you break this apart into seperate conditions, you're changing the functionality. It initially is essentially testing that a < b && b == c for your original statement of a < b == c.

Another example:

>>> 1 < 5 < 3
False

>>> (1 < 5) < 3
True

>>> 0 < 0 == 0
False

This is a chained comparison. It returns true if each pairwise comparison in turn is true. It is the equivalent to (0 < 0) and (0 == 0)

>>> (0) < (0 == 0)
True

This is equivalent to 0 < True which evaluates to True.

>>> (0 < 0) == 0
True

This is equivalent to False == 0 which evaluates to True.

>>> 0 < (0 == 0)
True

Equivalent to 0 < True which, as above, evaluates to True.

Looking at the disassembly (the bytes codes) it is obvious why 0 < 0 == 0 is False.

Here is an analysis of this expression:

>>>import dis

>>>def f():
...    0 < 0 == 0

>>>dis.dis(f)
  2      0 LOAD_CONST               1 (0)
         3 LOAD_CONST               1 (0)
         6 DUP_TOP
         7 ROT_THREE
         8 COMPARE_OP               0 (<)
        11 JUMP_IF_FALSE_OR_POP    23
        14 LOAD_CONST               1 (0)
        17 COMPARE_OP               2 (==)
        20 JUMP_FORWARD             2 (to 25)
   >>   23 ROT_TWO
        24 POP_TOP
   >>   25 POP_TOP
        26 LOAD_CONST               0 (None)
        29 RETURN_VALUE

Notice lines 0-8: These lines check if 0 < 0 which obviously returns False onto the python stack.

Now notice line 11: JUMP_IF_FALSE_OR_POP 23 This means that if 0 < 0 returns False perform a jump to line 23.

Now, 0 < 0 is False, so the jump is taken, which leaves the stack with a False which is the return value for the whole expression 0 < 0 == 0, even though the == 0 part isn't even checked.

So, to conclude, the answer is like said in other answers to this question. 0 < 0 == 0 has a special meaning. The compiler evaluates this to two terms: 0 < 0 and 0 == 0. As with any complex boolean expressions with and between them, if the first fails then the second one isn't even checked.

Hopes this enlightens things up a bit, and I really hope that the method I used to analyse this unexpected behavior will encourage others to try the same in the future.