Why does comparison of bytes with str fails in Python3

Question

In Python3 this expression evaluates as False:

b"" == ""

while in Python2 this comparison is True:

u"" == ""

Checking for identity with is obviously fails in both cases.

But why would they implement such a behaviour in Python3?

Answer 1

In Python 2.x, the design goal for unicode is to enable transparent operations between unicode & byte strings by implicitly converting between the 2 types. When you do the comparison u"" == "", the unicode LHS is automatically encoded into a byte string first, and then compared to the str RHS. That's why it returned True.

In contrast, Python 3.x, having learned from the mess of unicode that was in Python 2, decided to make everything about unicode vs. byte strings explicit. Thus, b"" == "" is False because the byte string is no longer automatically converted to unicode for comparison.

Answer 2

In python 3 string is Unicode . The type used to hold text is str and the type used to hold data is bytes.

the str and bytes types cannot be mixed, you must always explicitly convert between them. Use str.encode() to go from str to bytes, and bytes.decode() to go from bytes to str.

Therefore, if you do b"".decode() == "" you'll get True :

>>> b"".decode() == ""
True

For more info read Text Vs. Data Instead Of Unicode Vs. 8-bi

Answer 3

The designers decided to not assume an encoding for coercion when comparing bytes to strings, so it falls under the default behavior of Python 3.x whereby comparisons containing differing types fail.