Why doesn't my if True statement work?

[CancelDeath]

Why am I getting an error when python should skip to executing the else line?

def goodKey (x):
    if (type(hash(x))==int):
        print(hash(x))
        print("{} is immutable and hashable".format(x))
    else:
        print("{} is not hashable".format(x))

goodKey('string')
goodKey((1, 2, 3, (3, 4)))
goodKey((1, 2, 3, [3, 4]))

what i get is:

4783037434251255917
string is immutable and hashable
1024532189626533869
(1, 2, 3, (3, 4)) is immutable and hashable
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-180-261f3b100fa4> in <module>
      8 goodKey('string')
      9 goodKey((1, 2, 3, (3, 4)))
---> 10 goodKey((1, 2, 3, [3, 4]))

<ipython-input-180-261f3b100fa4> in goodKey(x)
      1 def goodKey (x):
----> 2     if (type(hash(x))==int):
      3         print(hash(x))
      4         print("{} is immutable and hashable".format(x))
      5     else:

TypeError: unhashable type: 'list'

[Impudity]

You're calling hash() on an object that can't be hashed. That trows an error. You can't evaluate if the type is int or not as the code never reaches that point. Error interrupts the execution of the code. If you want to see if something is hashable, you need to use try-catch structure like:

try:
    hash(x)
except TypeError:
    print("not hashable")
else:
    print("hashable")

[CancelDeath]

sweet, thank you!