Python Why does Python 3's print(1!=0==0) return 'True'?

[Bipolarity]

print(1!=0==0) seems to print the value 'True'. I'm trying to understand why.

Does it first evaluate (1!=0) ?

Or does it first evaluate (0==0) ?

Or does it separate them into (1!=0)^(0==0) ?

I appreciate all help.

BiP

 

[jedishrfu]

Not sure but I think it would use the left to right rule since != and == would have the same precedence as operators.

It's best to write code without an ambiguity like this by explicitly using parenthesis to order the sequence of operations.

 

[D H]

Or does it separate them into (1!=0)^(0==0)

That's what it does. There's some python comparison chaining magic going on here to enable expressions such as 0<x<1 to mean x is between 0 and 1 -- just like one would write in math. That you can use it for 1!=0==0 is a side effect of this magic.

 

[uart]

Or does it separate them into (1!=0)^(0==0) ?

That's what it does. There's some python comparison chaining magic going on here to enable expressions such as 0<x<1 to mean x is between 0 and 1 -- just like one would write in math. That you can use it for 1!=0==0 is a side effect of this magic.

Thanks for the tip on the 0<x<1 syntax DH. That's useful to know. :)

Though surely it's equivalent to (1!=0) and (0==0), rather than xor.

 

[D H]

Though surely it's equivalent to (1!=0) and (0==0), rather than xor.

Correct. I read the ^ as the mathematical shorthand for boolean and, which is not the case.It's not quite equivalent to (1!=0) and (0==0). There is a subtle difference. Consider

self.inbounds = 0 < self.some_function() < 1​

This will call self.some_function() once and only once. Now let's rewrite this expression as

self.inbounds = (0 < self.some_function()) and (self.some_function() < 1)​

With this rewrite, self.some_function() will be called once if the result from the first call is non-positive, twice if it is positive. Calling the function sometimes once, sometimes twice, can be deleterious if the function has side effects.