steven Michalske added the comment:
The RE compiler will not error out, with a back reference in there...
It treats the {\1} as a literal {\1} in the string.
In [180]: re.search("(\d) fo.{\1}", '3 foo{\1}').group(0)
Out[180]: '3 foo{\x01}'
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Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue20678>
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