On Wednesday, January 22, 2020 at 1:35:16 PM UTC-8, Simon King wrote:
>
> Here is an example:
> Unpickling the data initially results in a string, s:
> sage: s = '\x80\x1f'
> I want it to be interpreted as the following bytes, b:
> sage: b = b'\x80\x1f'
>
> How can I efficiently transform s into b? The following works, but I
>
This is exactly what we encountered on #28444, where we found that "latin1"
decoding has a left-inverse (if decoding acts on the left) of "latin1"
encoding. It looks like the encoding Py3 used for your py2-string was
"latin1" (that's what we decided was the smart thing to do most of the time
for possibly binary data), so
s.encode(encoding="latin1")
should do the trick. That should be pretty efficient.
> doubt that it is very efficient:
> sage: import struct
> sage: struct.pack('{}B'.format(len(s)),*(ord(_) for _ in s)) == b
> True
>
> Note that sage.cpython.string.str_to_bytes does't do what I need:
> sage: sage.cpython.string.str_to_bytes(s)
> b'\xc2\x80\x1f'
>
> Best regards,
> Simon
>
>
>
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