Python: Why does equality comparing an int with a string not throw an error?

Question

In Python 3 the attempt to order a string and an int (e.g. 1 > \"1\") throws a TypeError. Why does comparing a string with an int for equality not throw an e

Answer 1

The reason the orderings raise TypeError on non-comparable objects is to represent that there is no sensible answer, rather than any prediction on whether it would ever be useful. Allowing an equality test is sensible by this logic, insofaras there is an answer for "are two non-comparable objects equal?" (they aren't). See, eg, http://www.gossamer-threads.com/lists/python/dev/919516 .

Answer 2

This allows you, for example, to have a dictionary with keys of mixed types.

If you couldn't compare 1 and "1" for equality, you wouldn't be able to use them as keys into the same dictionary.

As things stand, you can compare them, and they always compare unequal:

The objects need not have the same type. If both are numbers, they are converted to a common type. Otherwise, objects of different types always compare unequal, and are ordered consistently but arbitrarily.

Answer 3

Strength and weakness of languages typing

Typing of language could be strong or weak (loose). The stronger typing language has the less different types could be operated in the same operation. Weakness and strength of language typing don't have exact threshold - some language could have stronger typing then other and weaker than another one. Python typing is much stronger than JS.

== implemented as more of less weakly-typed operation. It can compare different types but you need to have both values of the same type to have a chance to obtain True. a == b #true means a and b is objects of the same type and have the equal values. > < in Python 3 implemented as strongly-typed operation and couldn't be performed on different types.