问题
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?
回答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.
回答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. Usestr.encode()
to go from str to bytes, andbytes.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
回答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.
来源:https://stackoverflow.com/questions/30580386/why-does-comparison-of-bytes-with-str-fails-in-python3