Given 2 int values, return True if one is negative and other is positive

Question

def logical_xor(a, b): # for example, -1 and 1
    print (a < 0) # evaluates to True
    print (b < 0) # evaluates to False
    print (a < 0 != b < 0) # EVALUATES TO FALSE! why??? it's True != False
    return (a < 0 != b < 0) # returns False when it should return True

print ( logical_xor(-1, 1) ) # returns FALSE!

# now for clarification

print ( True != False) # PRINTS TRUE!

Could someone explain what is happening? I'm trying to make a one liner:

lambda a, b: (a < 0 != b < 0)

Answer 1

All comparison operators in Python have the same precedence. In addition, Python does chained comparisons. Thus,

(a < 0 != b < 0)

breaks down as:

(a < 0) and (0 != b) and (b < 0)

If any one of these is false, the total result of the expression will be False.

What you want to do is evaluate each condition separately, like so:

(a < 0) != (b < 0)

Other variants, from comments:

(a < 0) is not (b < 0) # True and False are singletons so identity-comparison works

(a < 0) ^ (b < 0) # bitwise-xor does too, as long as both sides are boolean

(a ^ b < 0) # or you could directly bitwise-xor the integers; 
            # the sign bit will only be set if your condition holds
            # this one fails when you mix ints and floats though

(a * b < 0) # perhaps most straightforward, just multiply them and check the sign